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2 years ago

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If the diameter of a circle changes from 5 cm to 15 cm, how will the circumference change? A) multiplies by 1 3 B) multiplies by

3 C) increases by 10 D) decreases by 10

2 answers:

andreyandreev [35.5K]2 years ago

5 0

The length of a circumference: l=πD
l₁=π×5=5π
l₂=π×15=15π
Then l₂÷l₁=15π÷5π=3 and the answer is B)

Debora [2.8K]2 years ago

4 0

Given that diameter of the old circle = 5 cm

Given that diameter of the new circle = 15 cm

Now we have to find the change in circumference from old to new circle.

To find that we need formula of circumference of the circle which is

circumference = D \pi

where D represents diameter of the circle.

Then ratio of the circumference of the both circle is given by

$\frac{Circumference_{old}}{Circumference_{new}}=\frac{D_{old} \pi}{D_{new} \pi}$

$\frac{Circumference_{old}}{Circumference_{new}}=\frac{5 \pi}{15 \pi}$

$\frac{Circumference_{old}}{Circumference_{new}}=\frac{1}{3 }$

That means circumference of the new circle is 3 times more than old circle.

Hence choice "B) multiplies by 3" is correct.

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Refer to the following scenario:You want to see if there is a difference between the exercise habits of Science majors and Math

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Answer:

1. H0: P1 = P2

2. Ha: P1 ≠ P2

3. pooled proportion p = 0.542

4. P-value = 0.0171

5. The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

6. The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

Step-by-step explanation:

We should perform a hypothesis test on the difference of proportions.

As we want to test if there is significant difference, the hypothesis are:

Null hypothesis: there is no significant difference between the proportions (p1-p2 = 0).

Alternative hypothesis: there is significant difference between the proportions (p1-p2 ≠ 0).

The sample 1 (science), of size n1=135 has a proportion of p1=0.607.

$p_1=X_1/n_1=82/135=0.607$

The sample 2 (math), of size n2=92 has a proportion of p2=0.446.

$p_2=X_2/n_2=41/92=0.446$

The difference between proportions is (p1-p2)=0.162.

$p_d=p_1-p_2=0.607-0.446=0.162$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{82+41}{135+92}=\dfrac{123}{227}=0.542$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.542*0.458}{135}+\dfrac{0.542*0.458}{92}}\\\\\\s_{p1-p2}=\sqrt{0.001839+0.002698}=\sqrt{0.004537}=0.067$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.162-0}{0.067}=\dfrac{0.162}{0.067}=2.4014$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$\text{P-value}=2\cdot P(z>2.4014)=0.0171$

As the P-value (0.0171) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

We want to calculate the bounds of a 99% confidence interval of the difference between proportions.

For a 99% CI, the critical value for z is z=2.576.

The margin of error is:

$MOE=z \cdot s_{p1-p2}=2.576\cdot 0.067=0.1735$

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

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.162-0.1735=-0.012\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.162+0.1735=0.335$

The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

6 0

2 years ago

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

melomori [17]

Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ $b+c=20$
Equation [2] ⇒ $1.50b+5c=54.50$

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; $b$ and $c$ , that we need to solve, then this becomes a simultaneous equation problem.

First, rearrange Equation [1] to make either $b$ or $c$ the subject:
$b+c=20$
$b=20-c$

Then we substitute $b=20-c$ into Equation [2]
$1.50b+5c=54.50$
$1.50(20-c)+5c=54.50$
$30-1.50c+5c=54.50$
$5c-1.5c=54.50-30$
$3.5c=24.50$
$c=7$

Now we know the value of $c$ which is $c=7$ , substitute this value into $b=20-c$ we have $b=20-7=13$

Answer:
Numbers of books = 13
Numbers of CDs = 7

5 0

2 years ago