El elemento neutro de la suma es el 1.

Answer:

10 liters of water

Step-by-step explanation:

Given

[tex]A = 10L\ milk[/tex]

Required

How much the second holds

Since they are both identical, then both containers hold the same volume (even if the quantity they hold are different)

Hence:

[tex]B = 10L\ water[/tex]

Answer:

42+123+x=180

x=180-165

x=15 degrees

Step-by-step explanation:

the test statistic (t = -12.857) is smaller than the critical value (-1.860), we have enough evidence to reject the null hypothesis.

To test the hypothesis regarding the population mean, we can perform a one-sample t-test.

Given:

- Null hypothesis (H₀): μ = 13

- Alternative hypothesis (Hₐ): μ < 13

- Sample mean ([tex]\bar{X}[/tex]) = 10

- Sample standard deviation (s) = 0.7

- Sample size (n) = 9

- Significance level (α) = 0.05

To conduct the t-test, we can calculate the test statistic and compare it with the critical value from the t-distribution.

The test statistic (t-score) is calculated as:

t = ([tex]\bar{X}[/tex] - μ) / (s / √n)

Plugging in the values:

t = (10 - 13) / (0.7 / √9)

t = -3 / (0.7 / 3)

t = -3 / 0.233

t ≈ -12.857

To determine the critical value, we need to find the appropriate degrees of freedom (df) for a one-sample t-test. In this case, df = n - 1 = 9 - 1 = 8.

Using a significance level of α = 0.05 and looking up the critical value for df = 8 in the t-distribution table, we find the critical value to be approximately -1.860.

Since the test statistic (t = -12.857) is smaller than the critical value (-1.860), we have enough evidence to reject the null hypothesis.

Decision: Based on the test results, at α = 0.05, we reject the null hypothesis (H₀: μ = 13). There is sufficient evidence to support the alternative hypothesis (Hₐ: μ < 13), suggesting that the population mean is less than 13.

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Given question is incomplete, the complete question is below

Assuming the sample was taken from a normal population, test at α = 0.05 and state the decision. Họ: μ = 13 HA: μ < 13 [tex]\bar{X}[/tex]= 10 s= 0.7 n = 9

Answer:

b = 6.1 ft

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the perpendicular height )

Here A = 10.98 and h = 3.6 , then

[tex]\frac{1}{2}[/tex] × b × 3.6 = 10.98

1.8b = 10.98 ( divide both sides by 1.8 )

b = 6.1

Answer:

65 girls and 45 boys

Step-by-step explanation:

Answer: 11x + 20y ≥ 220

x + y ≤ 15

Hope this helps

9514 1404 393

Answer:

Step-by-step explanation:

The two inequalities represent the two relations described in the problem statement.

  x + y ≤ 15 . . . . . . . Mia works a maximum of 15 hours

  11x + 20y ≥ 220 . . . . Mia makes at least $220

These are graphed in the attachment. The solution area is the doubly-shaded area with vertices at (9, 6), (0, 11) and (0, 15). One possible solution is shown at (3, 11), which represents ...

  3 hours babysitting

  11 hours lifeguarding . . . . . . . . total 14 hours for $253

Answer:

not a right triangle

Step-by-step explanation:

you can use the pythagorean theorem to prove whether not this is a right triangle

if this triangle is a right triangle then the following equality should be true

a²+b²=c²

(9.75)²+(3.45)²=(10.24)²

(95.06)+(11.90)=(104.86)

106.96≠104.86

since the following equality is not true, this is not a right triangle.

Answer:

12

Step-by-step explanation:

1/5 mint = 6

30 - 6 = 24

Half of 24 is 12

12 is chocolate

so 12 must be vanilla

Answer:

a. The probability of all girls in a family with four children is 1/16 or 0.0625.

b. The probability of all girls given there is at least one girl is 1/15 or 0.0667.

c. The probability of having at least one boy and one girl in a family with four children is 15/16 or 0.9375.

a. To calculate the probability of all girls, we need to consider the possible outcomes of each child being a girl. Since each child has an independent probability of being a girl or a boy, the probability of all girls is (1/2) * (1/2) * (1/2) * (1/2) = 1/16 or 0.0625.

b. Given that there is at least one girl, we have three remaining children. The probability of all girls among the three remaining children is (1/2) * (1/2) * (1/2) = 1/8. Therefore, the probability of all girls given there is at least one girl is 1/8 divided by the probability of having at least one girl, which is 1 - (1/2)⁴ = 15/16, resulting in a probability of 1/15 or approximately 0.0667.

c. The probability of having at least one boy and one girl can be calculated by subtracting the probability of having all boys from the total probability space. The probability of having all boys is (1/2)⁴ = 1/16. Therefore, the probability of having at least one boy and one girl is 1 - 1/16 = 15/16 or approximately 0.9375. This probability accounts for all possible combinations of boys and girls in a family with four children, excluding the scenario of having all boys.

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Answer:

y = 7x

Step-by-step explanation:

you mutiply 7 times 1=7

you just continue to do that

Answer: 6 liters

Step-by-step explanation:

24 liters

He sells 75%

24 x 0.75 = 18 liters

24 - 18 = 6

He still has 6 liters left

The corresponding z-scores for the given observations can be calculated using the formula:

[tex]\[z = \frac{{x - \mu}}{{\sigma}}\][/tex]

where [tex]\(x\)[/tex] is the observation, [tex]\(\mu\)[/tex] is the mean, and [tex]\(\sigma\)[/tex] is the standard deviation.

For the baby born in week 34 weighing 2875 grams:

[tex]\[z_{34} = \frac{{2875 - 2500}}{{600}} = 0.625\][/tex]

For the baby born in week 41 weighing 3175 grams:

[tex]\[z_{41} = \frac{{3175 - 2800}}{{415}} = 0.904\][/tex]

To determine which baby weighs more relative to the gestation period, we compare the z-scores.

A. The baby born in week 34 weighs relatively more since its z-score 0.625 is smaller than the z-score 0.904 for the baby born in week 41.

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Answer:

3

Step-by-step explanation:

a.  the value of P(2.4 < x < 3) is 8.1.

b. the value of E(X) is 10.

Given function is f(x) = 5(6)x for x≥1 and elsewhere for a continuous random variable z.  

a. Compute P(2.4 < x < 3)

For continuous random variable, P(a < x < b) = ∫f(x)dx, where f(x) is the probability density function (PDF).

Here, f(x) = 5(6)x for x≥1 and elsewhere

So, P(2.4 < x < 3) = ∫f(x)dx = ∫2.4^35(6)xdx= 5 ∫2.4^36xdx= 5 [(3^2 - 2.4^2)/2] = 5 [(9 - 5.76)/2] = 5 [1.62] = 8.1

Hence, the value of P(2.4 < x < 3) is 8.1.

b. Compute E(X)Expected value of X is given by E(X) = ∫xf(x)dxFor continuous random variable, E(X) = ∫xf(x)dx, where f(x) is the probability density function (PDF).Here, f(x) = 5(6)x for x≥1 and elsewhereSo, E(X) = ∫xf(x)dx = ∫1∞5(6)x.xdx+ ∫-∞0 0dx= 5 ∫1∞6x^2dx+ 0 = 5 [(6) (x^3)/3]1∞= 5 [(6) (1^3)/3] = 10

Hence, the value of E(X) is 10.

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Given f(x)= 5 (6) x for rexeh 1 o elsewhere for a continuous veidon variable z. We are required to compute p(2.4 < x <3) and E(x).

Compute p(2.4 < x <3)For the given function, f(x) = 5 (6) x for 1 ≤ x ≤ 3 and 0 elsewhere.

So, the total area under the curve will be equal to 5 (6) 2 = 60.

And the required probability is given by the area under the curve from 2.4 to 3.  [Illustration is provided below]

Hence, p(2.4 < x <3) = 3.6

(b)Compute E(x)Expected value of x, E(x) is given by E(x) = ∫xf(x) dx,

which is equal to the area under the curve multiplied by the distance over which the function is spread.

Let's calculate the area under the curve, which is equal to 60, as we have calculated earlier.

Now, we calculate the distance over which the function is spread.

Distance = 3 - 1 = 2 unitsHence, E(x) = 60/2 = 30.

Answer:Therefore, p(2.4 < x <3) = 3.6 and E(x) = 30.

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Add 4.5

---

hope it helps

Answer: add 4.5 to both sides

Step-by-step explanation: this will allow the 4.5 to be removed from the y side and since you added it to one side you have to add it to the other side. This will give your y the value of 16.7.

Brainliest plz

Answer:

okay then

Step-by-step explanation:

Answer: gtyrehretr

Step-by-step explanation:

Answer:

1/3 or 33%

Step-by-step explanation:

15 + 15 + 20 +25 is equal to 75

25/75 is 1/3

Answer:

1 by 3

Step-by-step explanation:

15 + 15 + 20 + 25 = 75

so,

again,

Answer:

I would need to see the different tables to answer this question.

Answer:

1733.33

Step-by-step explanation:

The volume of a pyramid is (length*width*height) divided by 3

Answer:

1733.33

Step-by-step explanation:

v = (lwh)/3

v = (20 x 20 x 13)/3 = 5200/3 = 1733.33

Answer: 9 x 6 x 6

Step-by-step explanation: 9 x 6=54  54 x 6 = 324 ft

Answer:

c. 322

Step-by-step explanation:

Surface area of cylinder = 2πr² + 2πrh.

SA in terms of π = π(2r² + 2rh) →

π(2(7)² + 2(7)(16)) →

π(2(49) + 2(112)) → π(98 + 224) → π(322)

The given values are: Ksp of CaSO4 = 4.93 × 10⁻⁵, Molarity of Na2SO4 = 0.250(m). Molar mass of CaSO4 = 136.14 g/mol. We can write the equation for the dissolution of CaSO4 in water as:CaSO4(s) ⇌ Ca²⁺(aq) + SO₄²⁻(aq). Let's consider that "x" grams of CaSO4 dissolves in "1 L" of 0.250 M Na2SO4. Since the CaSO4 dissolves according to the following equation:CaSO4(s) ⇌ Ca²⁺(aq) + SO₄²⁻(aq). The concentration of Ca²⁺ ions in the solution will be "x" moles / 1 L. The concentration of SO₄²⁻ ions in the solution will be "x" moles / 1 L. Since the concentration of Na2SO4 in the solution is 0.250 M or 0.250 moles / L, the concentration of SO₄²⁻ ions contributed by Na2SO4 will be (2 × 0.250) M or 0.500 M. In order to determine the value of "x" or the amount of CaSO4 that dissolves, we need to consider the equilibrium of the solution. The Ksp expression for the dissolution of CaSO4 can be written as: Ksp = [Ca²⁺][SO₄²⁻], Ksp = (x)(x) = x². As the dissociation is very small compared to the concentration of Na2SO4, we can consider "0.250" moles of Na2SO4 in "1 L" of the solution completely dissociated. Thus, the final concentrations of Ca²⁺ and SO₄²⁻ ions in the solution will be:[Ca²⁺] = x moles / L[SO₄²⁻] = (x + 0.500) moles /L. Therefore, we can write the expression for the ion product: IP = [Ca²⁺][SO₄²⁻]IP = (x)(x + 0.500)As the value of Ksp is equal to the IP, we can write the expression for Ksp as: Ksp = x² + 0.500xWe can substitute the value of Ksp as 4.93 × 10⁻⁵M:4.93 × 10⁻⁵ = x² + 0.500x. Solving for "x", we get the following quadratic equation: x² + 0.500x - 4.93 × 10⁻⁵ = 0. Solving for "x" using the quadratic formula: x = 0.00796 g/L. Therefore, the solubility of CaSO4(s) in 0.250 M Na2SO4 solution at 25°C is 0.00796 g/L.

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Answer:

4324eqW

Step-by-step explanation:

3 + 5 =7

Answer:

C

Step-by-step explanation:

4x +8 = 2x -8

4x = 2x - 16

2x = -16

x = -8

1. When a fair die is tossed the expected value E[X1] = 3.5 and the variance var(X1) = 35/12.

When a fair die is tossed, each of the six possible outcomes has an equal probability of 1/6. Let X1 denote the score obtained in the first toss.

To calculate the expected value E[X1], we find the sum of all possible values of X1 multiplied by their respective probabilities:

E[X1] = (1 * 1/6) + (2 * 1/6) + (3 * 1/6) + (4 * 1/6) + (5 * 1/6) + (6 * 1/6) = 3.5

To calculate the variance var(X1), we use the formula:

var(X1) = E[X1^2] - (E[X1])^2

First, we find E[X1^2] by taking the sum of the squares of all possible values of X1 multiplied by their respective probabilities:

E[X1^2] = (1^2 * 1/6) + (2^2 * 1/6) + (3^2 * 1/6) + (4^2 * 1/6) + (5^2 * 1/6) + (6^2 * 1/6) = 91/6

Substituting the values into the formula, we calculate var(X1):

var(X1) = (91/6) - (3.5)^2 = 35/12

Therefore, E[X1] = 3.5 and var(X1) = 35/12.

2. The probability distribution of Y = |X1 - X2| is tabulated as follows:

Y |X1 - X2| P(Y)

0 0 1/6

1 1 2/6

2 2 2/6

3 3 1/6

To calculate E[Y], we find the sum of all possible values of Y multiplied by their respective probabilities:

E[Y] = (0 * 1/6) + (1 * 2/6) + (2 * 2/6) + (3 * 1/6) = 1

Therefore, E[Y] = 1.

3. The statements E(Z^2) = E(Y^2) and Var(Z) = Var(Y) are false.

E(Z^2) and E(Y^2) represent the expected values of the squares of the random variables Z and Y, respectively. Since Z = X1 - X2 and Y = |X1 - X2|, the squares of Z and Y have different probability distributions, leading to different expected values.

Similarly, Var(Z) and Var(Y) represent the variances of Z and Y, respectively. Since Z and Y have different probability distributions, their variances will generally not be equal.

Therefore, E(Z^2) ≠ E(Y^2) and Var(Z) ≠ Var(Y).

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