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3 years ago

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An automobile manufacturer is preparing a shipment of cars and trucks on a cargo ship that can carry 21,600 Tons. The cars weigh

3.6 tons each and the trucks weigh 7.5 tons each. For a different shipment, cars are loaded first, and then trucks are loaded afterwards. Write an equation you could enter into a calculator or a spreadsheet tool to find the number of trucks that can be shipped if the number of cars is known. Let x=number of cars (input) and Let y=number of trucks (output)

1 answer:

inysia [295]3 years ago

3 0

Answer:

Number of trucks (y) that can be shipped if the number of cars (x) is known $=\frac{21,600-3.6x}{7.5}$

Step-by-step explanation:

A cargo ship can carry 21,600 Tons.

Let x denotes number of cars and y denotes number of trucks.

The cars weigh 3.6 tons each and the trucks weigh 7.5 tons each.

Therefore,

$3.6x+7.5y=21,600$

Now write y in terms of x.

$7.5y=21,600-3.6x\\y=\frac{21,600-3.6x}{7.5}$

Number of trucks (y) that can be shipped if the number of cars (x) is known $=\frac{21,600-3.6x}{7.5}$

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GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = <u>1.31</u>

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Step-by-step explanation:

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2 years ago

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soldi70 [24.7K]

Given:

In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle.

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