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SashulF [63]
3 years ago
14

The number of nails of a given length is norm ally distributed with a mean length of 5.00 in, and a standard deviation of 0.03 i

n. Find the number of nails in a bag of 120 that are over 5.03 in. long.
A. About 40 nails
B. About 30 nails
C. About 20 nails
D. About 10 nails
Mathematics
1 answer:
natta225 [31]3 years ago
6 0

Answer:

120·(1 - Ф((5.03 - 5)/0.03)) ≈ 20 nails

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The graph below shows a household’s budget. If the household spends $2,000 a month, which of the following statements is not tru
andreev551 [17]
The THIRD statement is not true.
$2000 (spends a month)
The 5% of $2000 is only $100 for food.

The first statement is true "The household spend $260 per month on the car".
13 % of $2000 = $260

The second statement is true "The household spend $600 per month for rent".
30% of $2000 = $600

The fourth statement is also true "The household spend $200 in savings each month"
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3 0
3 years ago
Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last
lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55

The degrees of freedom are df=599

df=n-1=600-1=599

The P-value for this test statistic is:

P-value=P(t

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0
3 years ago
Plz help !!!!!!!!!!!
TiliK225 [7]

Answer:

The answer is 9.

4 0
3 years ago
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Stephanie can make 12 baskets in 25 seconds. At this rate, how many baskets could she make in 150 seconds?
Vedmedyk [2.9K]

Answer:

72 in 150 seconds.

12/25=0.48

0.48 per second (not full baskets note)

.48x150= 72 baskets

5 0
2 years ago
Lorelei filled her 5-gallon jug with water. How many times could she fill her 2-quart canteen with water from the jug?
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There are 4 quarts in each gallon.
It would take 2 fills of the canteen to make one gallon.
2x5=10

Also, this image may be helpful.

7 0
3 years ago
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