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4 years ago

8

Kameryn types 75 words per minute and is just starting to write a term paper. Joe already has 510 words written and types at a s

peed of 60 words per minute. For what number of minutes will kameryn have more words typed than joe?

1 answer:

AnnZ [28]4 years ago

6 0

Answer:

Kameryn will have more words typed than Joe when the number of minutes exceeds 34.

Step-by-step explanation:

Let

x -----> the number of minutes

y ----> the total words typed

we know that

<em>Kameryn</em>

$y=75x$ -----> equation A

<em>Joe</em>

$y=60x+510$ -----> equation B

Solve the system of equations by substitution

Substitute equation A in equation B and solve for x

$75x=60x+510$

$75x-60x=510$

$15x=510$

$x=34\ minutes$

That means

For x=34 minutes

The amount of words written by Kameryn and Joe are the same.

therefore

For x > 34 minutes

Kameryn will have more words typed than Joe when the number of minutes exceeds 34.

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

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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.

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3 years ago

The given expression represents the area. Find the side length of the square.

Vesnalui [34]

9514 1404 393

Answer:

(4x +5)

Step-by-step explanation:

The square of a binomial is ...

(a +b)² = a² +2ab +b²

We assume that the given trinomial is a perfect square, so we can find the terms of the squared binomial by looking at the square roots of the first and last terms:

√(16x²) = 4x

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So, our first choice of side length for the square is ...

s = (4x +5) . . . . . side length of the square

__

Using the above identity, we can verify that the square of this gives the right area:

A = s² = (4x +5)² = (4x)² +2(4x)(5) +5²

A = 16x² +40x +25 . . . . . . as required

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3 years ago