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Andrei [34K]
3 years ago
13

Solve.

Mathematics
2 answers:
Sophie [7]3 years ago
5 0

Answer: He swam about 900 meters on those 3 weeks.

Paha777 [63]3 years ago
4 0

Answer:

Step-by-step explanation:if three weeks is equal to 21 than 0.86 times 21 equal 18.06 i believe

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Jasmine’s grandma gave her $7 to spend at a yogurt stand. The frozen yogurt and toppings cost $0.45 per ounce. She also wants to
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Answer:

Jasmine can afford 10 ounces of yogurt and toppings.

Step-by-step explanation:

$7 - $2.50 bottle of water

= $4.50

$4.50 divided by $0.45

= 10 ounces

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3 years ago
Have you ever been in the middle of telling a story and had to use hand gestures or even a drawing to make your point? In algebr
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2 years ago
) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa
asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u>p_1<u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u>p_2<u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no significant difference in seat belt use}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

                     T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers 16-24 years old who buckle up = 79%

\hat p_2 = sample proportion of drivers 25-69 years old who buckle up = \frac{924}{1100} = 84%

n_1 = sample of 16-24 years old drivers = 1000

n_2 = sample of 25-69 years old drivers = 1100

So, <u><em>test statistics</em></u>  =  \frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }  

                              =  -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that there is a difference in seat belt use.

4 0
3 years ago
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Step-by-step explanation: if correct please mark brainliest

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3 years ago
su ling started an art project with 1 yard of felt. she used to 2/6 yards on Tuesday and 3/6 yards on Wednesday. how much felt d
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