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2 years ago

15

Mrs.Schroeder, a school principal ,drove her own car for a class trip ,taking a different route . The principal's car can travel

32 miles per gallon of gas,and it used 12 gallons for the trip. How many miles did the principal drive?

2 answers:

eimsori [14]2 years ago

5 0

The principal drove 384 miles

dsp732 years ago

3 0

Answer:

the principal drove 384 miles

Step-by-step explanation:

12x32=384 miles

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Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.ca

Phoenix [80]

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p =0.09$ represent the estimated proportion for the sample

n=800 is the sample size required

$z$ represent the critical value for the margin of error

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error is given by :

$Me=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198$

And replacing into the confidence interval formula we got:

$0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702$

$0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098$

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

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1 year ago

A shoe factory produces 1.81 × 104 pairs of shoes each month. Estimate how many pairs of shoes the factory will produce if it ma

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Answer:

2*10 ^6 shoes in 10 years

Step-by-step explanation:

1.81 × 10^4 pairs of shoes each month

1.2 × 10^2 months in 10 years

Multiply these together to determine the number of shoes in 10 years

1.81 × 10^4 * 1.2 × 10^2

We want to estimate so 1.81 is approximately 2 and 1.2 is approximately 1

2 × 10^4 * 1 × 10^2

Multiply the numbers

2*1 = 2

Add the exponent

10 ^ ( 4+2)

10 ^6

Put this back together

2*10 ^6

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First let's write out the inequality before choosing a graph.

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y pounds of grapes: y

So...

1/3x + y < 5

The maximum weight is 4 pounds since the total weight of both the grapes and apples are less than 5.

In the y-axis, the first, third, and fourth graphs already exceed the capacity of 5 pounds.

So, by process of elimination, the correct graph for this problem is the second one.

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Answer:

Step-by-step explanation:

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