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Lynna [10]
2 years ago
12

Jace walked a total distance of StartFraction 5 Over 6 EndFraction of a mile to school this week. If he walked the same distance

each of the 5 days, how far did he walk each day? One-sixth of a mile One-fifth of a mile StartFraction 6 Over 25 EndFraction of a mile StartFraction 25 Over 6 EndFraction miles
Mathematics
2 answers:
geniusboy [140]2 years ago
4 0

Answer:

b

Step-by-step explanation:

serg [7]2 years ago
3 0

Answer:

b

Step-by-step explanation:

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Customer: "I see that the amount I owe has been reduced from $90.00 to $75.00." Employee: "Yes, the discount has saved you _____
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Jeremy earned $33,000 after working for 44 weeks. How much money did Jeremy earn each week? Show Your Work
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3 years ago
If a square has a perimeter of 36 in. what is the Circumstances of the circle?given the exact answer in terms of pi.
vfiekz [6]

Answer:

a)B)9π

b)G)16π in²

Step-by-step explanation:

according to the question

64 \times  \frac{\pi}{4}

16\pi \:  {in}^{2}

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C/d=π

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C=9π

6 0
3 years ago
A billing company that collects bills for​ doctors' offices in the area is concerned that the percentage of bills being paid by
omeli [17]

Answer:

1) A. H0: p = 0.30

HA: p not equal to 0.30

2) A. The Independence Assumption is met.

C. The Randomization Condition is met.

D. The Success/Failure Condition is met.

3) Test statistic z = 2.089

P-value = 0.0367

4) C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.

Step-by-step explanation:

1) This is a hypothesis test for a proportion.

The claim is that there is a significant change in the percent of bills being paid by medical​ insurance.

As we are looking for evidence of a difference, no matter if it is higher or lower than the null hypothesis proportion, the alternative hypothesis is defined by a unequal sign.

Then, the null and alternative hypothesis are:

H_0: \pi=0.3\\\\H_a:\pi\neq 0.3

2) Cheking the conditions:

The independence assumption and the randomization condition are met as the bills were selected randomly from the population.

The 10% condition can not be checked, as we do not know the size of the population.

The success/failure condition is met as the products np and n(1-p) are bigger than 10 (the number of successes and failures are both bigger than 10).

3) The significance level is assumed to be 0.05.

The sample has a size n=9260.

The sample proportion is p=0.31.

 

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.31-0.3-0.5/9260}{0.005}=\dfrac{0.01}{0.005}=2.089

This test is a two-tailed test, so the P-value for this test is calculated as:

\text{P-value}=2\cdot P(z>2.089)=0.0367

As the P-value (0.0184) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there is a significant change in the percent of bills being paid by medical​ insurance.

5 0
3 years ago
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