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3 years ago

7

describe a box a box that has a volume of 5 cubic units. Then describe a box that has 3 times the volume of the 1st box.what is

the volume of the second box

1 answer:

qwelly [4]3 years ago

4 0

Answer:

7

Step-by-step explanation:

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When Joseph first starts working at a grocery store, his hourly rate is \$10$10dollar sign, 10. For each year he works at the gr

morpeh [17]

Answer:

$R(t)=10+0.50t$

Step-by-step explanation:

Let t represent the number of years.

We have been given that When Joseph first starts working at a grocery store, his hourly rate is $10.

For each year he works at the grocery store, his hourly rate increases by $0.50. Increase in hourly rates after t years would be $0.50t$

The hourly rates after t years will be 10 plus $0.50t$ .

We can represent this information in an equation as:

$R(t)=10+0.50t$

Therefore, the function $R(t)=10+0.50t$ represents Joseph's hourly rates after t years.

3 0

3 years ago

A firm maintains 22 cars for business purposes. From past experience, it is known that approximately 10% will require major engi

seropon [69]

Answer:

The probability that at most 1 car will require major engine repair next year is P=0.3392.

Step-by-step explanation:

This can be modeled as a binomial random variable, with n=22 and p=0.1.

The probability that exavtly k cars will require major engine repair next year is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}$

Then, the probability that at most 1 car will require major engine repair next year is:

$P(x\leq1)=P(x=0)+P(x=1)\\\\\\P(x=0) = \dbinom{22}{0} p^{0}(1-p)^{22}=1*1*0.0985=0.0985\\\\\\P(x=1) = \dbinom{22}{1} p^{1}(1-p)^{21}=22*0.1*0.1094=0.2407\\\\\\P(x\leq1)=0.0985+0.2407\\\\P(x\leq1)=0.3392$

3 0

4 years ago

The value V of a certain automobile that is t years old can be modeled by V(t) = 14,651(0.81). According to the model, when will

lakkis [162]

Answer:

a) The car will be worth $8000 after 2.9 years.

b) The car will be worth $6000 after 4.2 years.

c) The car will be worth $1000 after 12.7 years.

Step-by-step explanation:

The value of the car after t years is given by:

$V(t) = 14651(0.81)^{t}$

According to the model, when will the car be worth V(t)?

We have to find t for the given value of V(t). So

$V(t) = 14651(0.81)^{t}$

$(0.81)^t = \frac{V(t)}{14651}$

$\log{(0.81)^{t}} = \log{(\frac{V(t)}{14651})}$

$t\log{(0.81)} = \log{(\frac{V(t)}{14651})}$

$t = \frac{\log{(\frac{V(t)}{14651})}}{\log{0.81}}$

(a) $8000

V(t) = 8000

$t = \frac{\log{(\frac{8000}{14651})}}{\log{0.81}} = 2.9$

The car will be worth $8000 after 2.9 years.

(b) $6000

V(t) = 6000

$t = \frac{\log{(\frac{6000}{14651})}}{\log{0.81}} = 4.2$

The car will be worth $6000 after 4.2 years.

(c) $1000

V(t) = 1000

$t = \frac{\log{(\frac{1000}{14651})}}{\log{0.81}} = 12.7$

The car will be worth $1000 after 12.7 years.

5 0

3 years ago

30% of the library books in the fiction section are worn and need replacement. 10% of the nonfiction holdings are worn. The libr

Nady [450]

Answer:

0.0048 or 0.48%

Step-by-step explanation:

Please see the attached files for details.

8 0

4 years ago

Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300.

Fynjy0 [20]

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10$ .

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when $X = 500$ . So

$Z = \frac{X - \mu}{s}$

$Z = \frac{500 - 465}{10}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.

8 0

3 years ago