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gladu [14]
2 years ago
6

I need some help again

Mathematics
1 answer:
Marrrta [24]2 years ago
4 0

A = small box of oranges

B = large box of oranges

3A + 14B = 203

11A + 11B = 220

Take the second equation and divide it by 11 as both A and B have the same coefficient, which will equal:

\frac{11A+11B}{11} = A + B

220 will also have to be divided by 11, giving you 20

A + B = 20

You can cancel out 3A in the first equation by subtracting 3A + 3B = 60 from it.

11B = 143

143/11 = 13

B (large box of oranges) costs $13.

Substitute 13 for B back into one of our equations, in this case 3A + 3B = 60

3A + 3*13 = 60

3A + 39 = 60

3A = 21

A = 7

Thus:

A small box of oranges costs $7, and a large box of oranges costs $13.

Hope this helped! (please brainliest)

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Answer:

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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.

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3 years ago
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GuDViN [60]

Answer:

The standard deviation of a probability distribution is a measure of variability of the distribution.

Step-by-step explanation:

We have been given an incomplete statement. We are asked to complete the given statement.

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It tells up how much a data point is spread out from the average or mean of the data set.

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Kaylis [27]

Given :

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To Find :

Two possible lengths for CD.

Solution :

Their are two cases :

1)

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C                  D          E

Here, CD = CE - DE

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CD = 12.3 cm

2)

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D      E              C

Here, CD = CE + DE

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Hence, this is the required solution.

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