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3 years ago

the density of pure gold 1,204 pounds per cubic foot? how many cubic feet is 481.6lbs of pure gold?

Mathematics

1 answer:

Nadusha1986 [10]3 years ago

5 0

Answer:

Volume of Gold = 0.4 cubic feet.

Step-by-step explanation:

First we have to make sure the units are same as the units the answer is required in.

Answer should be given in cubic feet.

Density is given in pounds per cubic foot. And Mass is given in pounds. So units are in order. Therefore, we can go ahead with the calculation.

The Density of Gold = [Mass of Gold/Volume of Gold]

By substituting values,

$1204$ ponds per cubic foot = [ $481.6$ pounds / Volume of gold]

Making Volume of God the subject of the equation,

Volume of Gold = [ $481.6$ pounds / tex]1204[/tex] ponds per cubic foot ]

Volume of Gold = 0.4 cubic feet.

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8.75 is 0.5% of what number

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Answer:

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Step-by-step explanation:

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and it would be 0.5% of 8.75 is what.

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3 years ago

Jared has one pizza that has 12 slices. he wants to share his pizza with his two brothers. how many slices will each boy have if

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2 years ago

Read 2 more answers

In a simple random sample of 14001400 young people, 9090% had earned a high school diploma. Complete parts a through d below.

ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has increased.

Step-by-step explanation:

Let p = proportion of young people who had earned a high school diploma.

A sample of n = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

$\hat p=0.90$

(a)

The standard error for the estimate of the percentage of all young people who earned a high school diploma is given by:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Compute the standard error value as follows:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=\sqrt{\frac{0.90(1-0.90)}{1400}}\\$

$=0.008$

Thus, the standard error for the estimate of the percentage of all young people who earned a high school diploma is 0.0080.

(b)

The margin of error for (1 - α)% confidence interval for population proportion is:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

Compute the critical value of z for 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

$=1.96\times 0.0080\\=0.01568\\\approx1.6\%$

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

$CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)$

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has increased, the hypothesis is defined as:

H₀: The percentage of young people who earn high school diplomas has not increased, i.e. p = 0.80.

Hₐ: The percentage of young people who earn high school diplomas has not increased, i.e. p > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of p, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has increased.

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3 years ago

the total cost of a new autombile, including a 4% sales tax on the price of the automobile, is $28,835. how much of the total co

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3 years ago

Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118. If a recent test-taker

LuckyWell [14K]

Answer:

Probability that the student scored between 455 and 573 on the exam is 0.38292.

Step-by-step explanation:

We are given that Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118.

Let X = Math scores on the SAT exam

So, X ~ Normal( $\mu=514,\sigma^{2} =118^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean score = 514

$\sigma$ = standard deviation = 118

Now, the probability that the student scored between 455 and 573 on the exam is given by = P(455 < X < 573)

P(455 < X < 573) = P(X < 573) - P(X $\leq$ 455)

P(X < 573) = P( $\frac{X-\mu}{\sigma}$ < $\frac{573-514}{118}$ ) = P(Z < 0.50) = 0.69146

P(X $\leq$ 2.9) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{455-514}{118}$ ) = P(Z $\leq$ -0.50) = 1 - P(Z < 0.50)

= 1 - 0.69146 = 0.30854

The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.69146.

Therefore, P(455 < X < 573) = 0.69146 - 0.30854 = 0.38292

Hence, probability that the student scored between 455 and 573 on the exam is 0.38292.

7 0

3 years ago