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3 years ago

9

In this diagram, ABAC – AEDF. If the

1 answer:

Volgvan3 years ago

6 0

Answer:

48/5in2=9.6in2 ~ 10in2

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Mitch is making a target for the beanbag toss at his school's fall fun-run and carnival. the target is a short round basket divi

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3/4 is your fraction because 3 out of the four sections gives you the bear

Step-by-step explanation:

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2 years ago

The scores of fourth grade students on a mathematics achievement test follow a normal distribution with a mean of 75 and standar

mamaluj [8]

Answer:

The appropriate solution is:

(a) 0.1056

(b) 0

(c) 0.9544

Step-by-step explanation:

The given values are:

Mean,

$\mu = 75$

Standard deviation,

$\sigma = 4$

(a)

⇒ $P(x>80)=1-(x$

$=1-P[\frac{x-\mu}{\sigma}$

$=1-P[\frac{x-\mu}{\sigma}$

$=1-P(z$

By using the table, we get

$=1-0.8944$

$=0.1056$

(b)

According to the question, the values are:

$n$ = 64

$\mu_\bar{x}$ = 75

Now,

⇒ $\sigma_\bar{x}$ = $\frac{\sigma}{\sqrt{n} }$

= $\frac{4}{\sqrt{64} }$

= $\frac{4}{8}$

= $0.5$

⇒ $P(\bar {x} >80 ) = 1 - P(\bar x$

$=1 - P[\frac{(\bar x-\mu_\bar x)}{\sigma \bar x} < \frac{80-75}{0.5} ]$

$=1-P(z$

By using the table, we get

$=1-1$

$=0$

(c)

As we know,

⇒ $\sigma_\bar x$ = $\frac{\sigma}{\sqrt{n} }$

= $\frac{4}{\sqrt{64} }$

= $\frac{4}{8}$

= $0.5$

then,

= $P(74< \bar x$

= $P[\frac{74-75}{0.5} < \frac{\bar x-\mu \bar x}{\sigma \bar x} < \frac{76-75}{0.5} ]$

= $P(-2$

= $P(z$

By using the table, we get

= $0.9772-0.0228$

= $0.9544$

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3 years ago

PLEASE HELP ME!! Simplify by multiplying the LCD

Umnica [9.8K]

Answer:

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3 years ago

I NEED HELP SO CONFUSED

nikklg [1K]

Answer:

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Step-by-step explanation:

nah jk its 69

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3 years ago

Read 2 more answers

10 10 x 10 10 x 10x10 x10

Sever21 [200]

The answer is 10,000,000 because u have to do 10x10x10x10x10x10x10

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2 years ago