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3 years ago

in kurt's school, the ratio of boys to girls is 4/3. of 120 students are boys, how many students are girls?

Mathematics

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

90 of the students are girls.

Step-by-step explanation:

Boys : Girls

B : G

4 : 3

120 : ?

(120 ÷ 4 = 30) so 4 × 30 = 120

(Do this to the other side) 3 × 30 = 90

B : G

120 : 90 [90 girls out of 210 total students]

Hope this helps. :)

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A chef uses a circular pan with a 24-inch diameter to make a pizza. What is the area of pizza

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Answer:

<u>The area of a pizza made in that circular pan is 452.39 inches²</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Diameter of the circular pan = 24 inches

Radius of the circular pan = 24/2 = 12 inches

2. What is the area of the pizza?

Let's use the formula of the area of a circle, this way:

Area = π * r²

Substituting with the real values, we have:

Area = 3.1416 * 12²

Area = 452.39 inches²

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Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was

Vika [28.1K]

Answer:

a) The 95% confidence interval for the mean is (-1.171, 11.717).

b) No, there is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure (at a 5% significance level).

Step-by-step explanation:

The data for both groups are:

Group 1 (calcium): 7 -4 18 17 -3 -5 1 10 11 -2

Group 2 (placebo): -1 12 -1 -3 3 -5 5 2 -11 -1 -3

The mean and standard deviation for Group 1 is:

$M_1=\dfrac{1}{10}\sum_{i=1}^{10}(7+(-4)+18+17+(-3)+(-5)+1+10+11+(-2))\\\\\\ M_1=\dfrac{50}{10}=5$

$s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\$

$s_1=\sqrt{\dfrac{1}{9}\cdot [(7-(5))^2+(-4-(5))^2+...+(-2-(5))^2]}\\\\\\$

$s_1=\sqrt{\dfrac{1}{9}\cdot [(4)+(81)+(169)+(144)+(64)+(100)+(16)+(25)+(36)+(49)]}\\\\\\s_1=\sqrt{\dfrac{688}{9}}=\sqrt{76.44}\\\\\\s_1=8.743$

The mean and standard deviation for Group 2 is:

$M_2=\dfrac{1}{11}\sum_{i=1}^{11}((-1)+12+(-1)+(-3)+3+(-5)+5+2+(-11)+(-1)+(-3))\\\\\\ M_2=\dfrac{-3}{11}=-0.273$

$s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{11}(x_i-M)^2}\\\\\\$

$s_2=\sqrt{\dfrac{1}{10}\cdot [(-1-(-0.273))^2+(12-(-0.273))^2+...+(-3-(-0.273))^2]}\\\\\\$

$s_2=\sqrt{\dfrac{1}{10}\cdot [(0.53)+(150.62)+...+(115.07)+(0.53)+(7.44)]}\\\\\\s_2=\sqrt{\dfrac{348.182}{10}}=\sqrt{34.82}\\\\\\s_2=5.901$

a) We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The Group 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The Group 2, of size n2=11 has a mean of -0.273 and a standard deviation of 5.901.

The difference between sample means is Md=5.273.

$M_d=M_1-M_2=5-(-0.273)=5.273$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288$

The degrees of freedom for this test are:

$df=n_1+n_2-1=10+11-2=19$

The critical t-value for a 95% confidence interval and 19 degrees of freedom is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.96 \cdot 3.288=6.444$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 5.273-6.444=-1.171\\\\UL=M_d+t \cdot s_{M_d} = 5.273+6.444=11.717$

The 95% confidence interval for the mean is (-1.171, 11.717).

b) This is a hypothesis test for the difference between populations means.

The claim is that increasing the amount of calcium in our diet reduce blood pressure.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

The significance level is 0.05.

The sample 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The sample 1, of size n1=11 has a mean of -0.273 and a standard deviation of 5.901.

The difference between sample means is Md=5.273.

$M_d=M_1-M_2=5-(-0.273)=5.273$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{5.273-0}{3.288}=\dfrac{5.273}{3.288}=1.604$

The degrees of freedom for this test are:

$df=n_1+n_2-1=10+11-2=19$

This test is a right-tailed test, with 19 degrees of freedom and t=1.604, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>1.604)=0.063$

As the P-value (0.063) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure.

4 0

3 years ago

Can any please help me with this god bless!

Katen [24]

Answer:

What do you need help with?

Step-by-step explanation:

8 0

3 years ago