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3 years ago

∛Which expression is NOT equivalent to this expression? 8(11y - 5) *

Mathematics

2 answers:

iren2701 [21]3 years ago

7 0

C
All other three if you distribute and multiply would give the same answer

Marina86 [1]3 years ago

3 0

You answer is A) 88y-40

Are you handing out brainliest. Promise it is right btw I used it on my test.

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natta225 [31]

Y=90
X=41
...........

4 0

3 years ago

Sharon called 3 friends on Friday. On Saturday, each of those 3 friends called 3 different friends. On Sunday, each person who w

Zarrin [17]

There was 27 people called on Sunday because its
3 x 3 = 9 for Saturday
then 9 x 3 = 27 for Sunday

6 0

4 years ago

Read 2 more answers

Use linear approximation, i.e. the tangent line, to approximate √ 16.2 as follows: Let f ( x ) = √ x . Find the equation of the

Dvinal [7]

The linear approximation to $f(x)$ near $x=16$ is

$L(x)=f(16)+f'(16)(x-16)$

We have

$f(x)=\sqrt x\implies f(16)=4$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(16)=\dfrac18$

$\implies L(x)=4+\dfrac18(x-16)=\dfrac x8+2$

Then

$f(16.2)\approx L(16.2)\iff\sqrt{16.2}\approx\dfrac{16.2}8+2$

$\implies\boxed{\sqrt{16.2}\approx4.025=\dfrac{161}{40}}$

5 0

3 years ago

Each person in a large square occupies, on average, 2.5 sq. feet of space. Estimate the number of people that can comfortably st

Aleksandr [31]

Answer:

2000 people

Step-by-step explanation:

The area of the gym is the width times length:

A = wl

A = (50 ft) (100 ft)

A = 5000 ft²

Write a proportion:

1 person / 2.5 ft² = x / 5000 ft²

x = 2000 people

3 0

3 years ago

Read 2 more answers

Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu

Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of $P(Z$ , you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0

4 years ago