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3 years ago

I forgot how to do smh. please help

Mathematics

1 answer:

Mazyrski [523]3 years ago

4 0

$\bf ~\hspace{7em}\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
~\hspace{4.5em}
a^n\implies \cfrac{1}{a^{-n}}
~\hspace{4.5em}
\cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{y^3z^5}{y^4z}\implies \cfrac{y^3z^5}{y^4z^1}\implies \cfrac{z^5\cdot z^{-1}}{y^{-3}y^4}\implies \cfrac{z^{5-1}}{y^{-3+4}}\implies \cfrac{z^4}{y}$

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Kim bought a poster that cost $8.95 and a ome colored pencils.The total cost was $21.35.How much did the colored pencils cost?

Dafna11 [192]

If the total cost was 21.35, and the poster cost 8.95, then all you need to do is subtract to poster cost from the total cost to find the cost of the pencil.
21.35-8.95= $12.40

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3 years ago

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Create two real world problems that involve the fraction 1/2 and the whole number 3

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You order 4 pizzas for a party. At the end of the party only half of a pizza is left. How much pizza did the guests eat? That's one problem can't think of a 2nd one.

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3 years ago

Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg

Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let X = number of college women on a diet.

The probability of a woman being on diet is, P (X) = p = 0.70.

The sample of women selected is, n = 267.

The random variable thus follows a Binomial distribution with parameters n = 267 and p = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions ( $\hat p$ ) follows a Normal distribution.

The mean of this distribution is:

$\mu_{\hat p} = p = 0.70$

The standard deviation of this distribution is: $\sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028$

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

$P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z$

**Use the z-table for the probability.

$P(\hat p \geq 0.75)=1-P(Z$

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

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3 years ago

What percentage of this shape is shaded?

Lilit [14]

Answer:

50%

Step-by-step explanation:

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3 years ago

alison, beatrice and chole each had some books. alison gave beatrice and chole some books that doubled the number of books they

slamgirl [31]

alison, beatrice and chole each of them have at first 56, 8, 32 books.

alison, beatrice and chole each had some books. alison gave beatrice and chole some books that doubled the number of books they had, lastly chole gave alison and beatrice some books that doubled the number of books they had. each of them had 32 books at the end.

<h3>What is Working backward?</h3>

Working backward is process in which calculation done in reverse direction.

let alison, beatrice and chole has x,y and z books initially,

alison gave beatrice and chole some books that doubled the number of books they had
alison, beatrice and chole has (96-2y-2z), (2y) , (2z)
chole gave alison and beatrice some books that doubled the number of books they had. each of them had 32 books at the end.

alison, beatrice and chole has 2(96-2y-2z), 2(2y) , (96-2(96-2y-2z)-4y)

at the end,
beatrice = 32

4y=32

y = 8

chole =32

96-2(96-2y-2z)-4y)=32
z=32

alison = 96-y-z

x=96-8-32
x=56

Thus the required alison, beatrice and chole each of them have at first 56, 8, 32 books.

Learn more about working backward here:

brainly.com/question/14636831

#SPJ1

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2 years ago