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Leni [432]
3 years ago
6

Henry divided his socks into five equal groups. Let s represent the total number of socks. Which expression and solution represe

nt the number of socks in each group if s = 20?
StartFraction s Over 5 EndFraction; when s = 20, the number of socks in each group is 4.
StartFraction s Over 5 EndFraction; when s = 20, the number of socks in each group is 15.
s minus 5; when s = 20, the number of socks in each group is 4.
s minus 5; when s = 20, the number of socks in each group is 15.
Mathematics
2 answers:
Scorpion4ik [409]3 years ago
4 0

Answer:

answer is A

Step-by-step explanation:

NNADVOKAT [17]3 years ago
4 0

Answer:

Correct ^^ (person above)

Step-by-step explanation:

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Christopher is trying to make extra money on his summer vacation from school. He
Anettt [7]

Answer:

The earnings from 6 hours washing cars and 5 hours landscaping is $114.

Step-by-step explanation:

Earnings = 9w + 12l

Earnings = 9(6)+12(5)=54+60=114

8 0
3 years ago
The breaking strengths of cables produced by a certain manufacturer have a mean, , of pounds, and a standard deviation of pounds
olchik [2.2K]

The question is incomplete. The complete question is :

The breaking strengths of cables produced by a certain manufacturer have a mean of 1900 pounds, and a standard deviation of 65 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 150 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1902 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased?

Solution :

Given data :

Mean, μ = 1900

Standard deviation, σ = 65

Sample size, n = 150

Sample mean, $\overline x$ = 1902

Level of significance = 0.01

The hypothesis are :

$H_0 : \mu = 1900$

$H_1 : \mu > 1900$

Test statics :

We use the z test as the sample size is large and we know the population standard deviation.

$z=\frac{\overline x - \mu}{\sigma / \sqrt{n}}$

$z=\frac{1902-1900}{65 / \sqrt{150}}$

$z=\frac{2}{5.30723}$

$z=0.38$

Finding the p-value:

P-value = P(Z > z)

             = P(Z > 0.38)

             = 1 - P(Z < 0.38)

From the z table. we get

P(Z < 0.38) = 0.6480

Therefore,

P-value = 1 - P(Z < 0.38)

            = 1 - 0.6480

             = 0.3520

Decision :

If the p value is less than 0.01, then we reject the H_0, otherwise we fail to reject  H_0.

Since the value of p = 0.3520 > 0.01, the level of significance, then we fail to reject  H_0.

Conclusion :

At a significance level of 0.01, we have no sufficient evidence to support that the mean breaking strength has increased.

4 0
2 years ago
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Katarina [22]

I dont know the answer

Step-by-step explanation:

little girl I am sorry

8 0
3 years ago
Whats 3,684,266 to the nearest hundred thousand and million
grin007 [14]
The nearest hundred thousand is 3,700,000
And the nearest million is 4,000,00
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3 years ago
What percentage of Coach passengers did not check bags
Andreyy89

Answer:

7

Step-by-step explanation:

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