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3 years ago

Square root of y^2 times y^2

Mathematics

1 answer:

Galina-37 [17]3 years ago

8 0

<span>Square root of y^2 = y
so
</span><span>Square root of y^2 times y^2 = y * y^2 = y^3

answer
y^3</span>

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Prove With Diagram Which fraction is larger, 5/8 or 3/4 Brainliest

4vir4ik [10]

Answer:

3/4

Step-by-step explanation:

Convert the fractions into decimals.

5/8 = 0.625

3/4 = 0.75

0.75 is greater than 0.625.

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2 years ago

What is an equivalent expression to 5(m+1)-1

adelina 88 [10]

Answer:

is this answer

Step-by-step explanation:

5(m+1)-1

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1 year ago

What is the factored form of the equation 2(x-1)(8x-3)=0? Which ordered pairs are solutions to the system? Check all of the boxe

vovikov84 [41]

Answer:

The solutions to the equation are 1, 0.375.

Step-by-step explanation:

The equation is already in factored form.

The solutions to the system are the values of x that make one or the other of the factors be zero.

x -1 = 0 . . . when x = 1

8x -3 = 0 . . . when x = 3/8 = 0.375

There is only one variable, so no "ordered pairs" are involved.

The solutions are {1, 0.375}.

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2 years ago

Read 2 more answers

The sum of a number y and -3 is -8 .

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2 years ago

The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assu

m_a_m_a [10]

Answer: 0.0170

Step-by-step explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e. $\mu=23.50$

$\sigma=5$

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let $\overline{x}$ be the sample mean.

Formula : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Then, the probability that the sample mean is at least $25.00 will be :-

$P(\overline{x}\geq\25.00)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\geq\dfrac{25-23.50}{\dfrac{5}{\sqrt{50}}})\\\\=P(z\geq2.12)\\\\=1-P(z$

Hence, the likelihood the sample mean is at least $25.00= 0.0170

5 0

2 years ago