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3 years ago

Jeff has a density of 1000 g/mL and a volume of 25-mL. What is Jeff’s mass?

Mathematics

1 answer:

zlopas [31]3 years ago

6 0

Answer:

25,000 grams

Step-by-step explanation:

Density can be found using the following formula:

$d=\frac{m}{v}$

If we rearrange the formula for $m$ , mass, we will multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass can be found by multiplying the density by the volume.

Jeff's density is 1000 grams per milliliter and his volume is 25 milliliters.

$d= 1000 g/mL\\v=25 mL$

Substitute the values into the formula.

$m=d*v$

$m= 1000 g/mL * 25 mL$

Multiply. When you multiply, the "mL" or milliliters will cancel out.

$m= 1000 g * 25$

$m= 25000 g$

Jeff's mass is 25,000 grams.

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3 years ago

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

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(b) The value of $P(\bar X$ is 0.7642.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$

And the standard deviation of the distribution of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74$

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(b)

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$P(\bar X$

$=P(Z$

*Use a z-table for the probability.

Thus, the value of $P(\bar X$ is 0.7642.

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Compute the value of $P(\bar X\geq 69.1)$ as follows:

Apply continuity correction as follows:

$P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)$

$=P(\bar X>69.6)$

$=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})$

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Thus, the value of $P(\bar X\geq 69.1)$ is 0.3670.

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