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2 years ago

Please help! (listing BRAINLIST and giving points) :)

Mathematics

2 answers:

Firlakuza [10]2 years ago

7 0

Answer:

7 hrs

Step-by-step explanation:

$187 for 11 hrs

$119 for x hrs

write a proportion.

187/11 = 119/x since Reuben is making money at the same rate.

solve for x:

187x = 1309

x=7

7 hours

joja [24]2 years ago

3 0

Answer:

7 hours

Step-by-step explanation:

Reuben works for 11 hours to earn $187

This means, his earnings rate is $187 per 11 hours of work

In 1 hour he would earn = 187/ 11 = $17

Therefore, he earns $17 in 1 hour

To find out how many hours he has to work to make $119, we multiply this by the rate.

He will earn $119 in = $119 x (1 hour/ $17) = 7 hours

Reuben has to work 7 hours to make $119

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3 years ago

Read 2 more answers

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages ar

Elenna [48]

Answer:

We conclude that the mean amount packaged is equal to 8.17 ounces.

Step-by-step explanation:

We are given that in a particular sample of 50 packages, the mean amount dispensed is 8.171 ounces, with a sample standard deviation of 0.052 ounces.

Let $\mu$ = population mean amount packaged.

So, Null Hypothesis, $H_0$ : $\mu$ = 8.17 ounces {means that the mean amount packaged is equal to 8.17 ounces}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 8.17 ounces {means that the mean amount packaged is different from 8.17 ounces}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount dispensed = 8.171 ounces

s = sample standard deviation = 0.052 ounces

n = sample of packages = 50

So, the test statistics = $\frac{8.171-8.17}{\frac{0.052}{\sqrt{50} } }$ ~ $t_4_9$

= 0.1359

The value of t-test statistics is 0.1359.

Also, the P-value of test-statistics is given by;

the meaning of the p-value is that the p-value is the probability of obtaining a sample mean that is equal to or more extreme than 0.001 ounces away from8.17 if the null hypothesis is true.

P-value = P( $t_4_9$ > 0.136) = More than 40% {from the t-table}

Since the P-value of our test statistics is more than the level of significance of 0.01, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean amount packaged is equal to 8.17 ounces.

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2 years ago

Please help! (listing BRAINLIST and giving points) ​:)​

Please help! (listing BRAINLIST and giving points) :)