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2 years ago

Question 11 of 37 What is the volume of the triangular prism shown below? 15 9

Mathematics

1 answer:

AVprozaik [17]2 years ago

8 0

Answer:

the correct answer is B. 135 cu. units

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What is 17/16 in simplest form

Olin [163]

It can be written as 1.0625

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3 years ago

Seven plus seven equal to

vazorg [7]

Answer:

the answer is 14

Step-by-step explanation:

7+7=14

I hope this help

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2 years ago

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Where would you insert parentheses to make the statement true?

otez555 [7]

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3 years ago

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The average college lecture hall (auditorium) can seat 60 students with a standard deviation of 21. Assume that a total of 60 le

fiasKO [112]

Answer:

The standard deviation of the sample mean is $\sigma _ {\= x } = 2.711$

Step-by-step explanation:

From the question we are told that

The mean is $\= x = 60$

The standard deviation is $\sigma = 21$

The sample size is $n = 60$

Generally the standard deviation of the sample mean is mathematically represented as

$\sigma _ {\= x } = \frac{\sigma }{\sqrt{n} }$

substituting values

$\sigma _ {\= x } = \frac{ 21 }{\sqrt{60} }$

$\sigma _ {\= x } = 2.711$

4 0

3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago