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rodikova [14]
3 years ago
15

One more geometry question! Need as soon as possible!

Mathematics
1 answer:
user100 [1]3 years ago
7 0

There's not enough information to find out how long segment CB is or any other segment for that matter.

Sorry I couldn't be of much help.

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Micheal bought a beautiful bouquet of flowers for $62.75 (not including tax). If he bought them while there was a 30% off specia
Fudgin [204]

Answer:

$81.575, or rounded, $81.58

Step-by-step explanation:

Since he got 30% off, you can just multiply by 1.30, or 130%, to get the original price.

3 0
3 years ago
Read 2 more answers
Eight times five plus nine
Vedmedyk [2.9K]

{49} 8•5=40. 9+40= 49

6 0
4 years ago
Suppose you pay $24 for a pair of shoes that has been discounted 20%. What is the original price of the shoes?
dalvyx [7]
The original price of the shoes was "x", which is the 100% price of the pair of shoes, but you arrived at the store when the 20% sale was going on, so you've got them for 100-20%, which we know is $24, but how much is "x" then?

well      \bf \begin{array}{ccllll}
amount&\%\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
x&100\\
24&100-20
\end{array}\implies \cfrac{x}{24}=\cfrac{100}{100-20}

solve for "x"
6 0
4 years ago
Barrons reported that the average number of weeks an individual is unemployed is 17.5 week. assume that for the population of al
slega [8]

Answer:

\mu = 17.5

\sigma = 4

n = 50

A) show the sampling distribution of x, the sample mean average for a sample of 50 unemployment individuals.

We will use central limit theorem

So, mean of sampling distribution = \mu_{\bar{x}}=\mu = 17.5

Standard deviation of sampling distribution = \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}= 0.5656

B) What is the probability that a simple random sample of 50 unemployment individuals will provide a sample mean within one week of the population mean?

A sample mean within one week of the population mean means x-\mu = 1

So, P(|x-\mu|

=P(\frac{-1}{\frac{4}{\sqrt{50}}}

=P(-1.77

=P(z

=0.9616-0.0384

=0.9232

The probability that a simple random sample of 50 unemployment individuals will provide a sample mean within one week of the population mean is 0.9232.

C) What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within a half week of the population mean?

A sample mean within one week of the population mean means x-\mu = 1

So, P(|x-\mu|

=P(\frac{-1-0.5}{\frac{4}{\sqrt{50}}}

=P(-0.88

=P(z

=0.8106-0.1894

=0.6212

The probability that a simple random sample of 50 unemployed individuals will provide a sample mean within a half week of the population mean is 0.6212

3 0
4 years ago
A) Complete the table of values for x + y = 6<br> Х<br> -2<br> -1<br> 0<br> 1<br> 2<br> 3<br> y
-BARSIC- [3]

Answer:

8 , 7 , 6 , 5 , 4 , 3

Step-by-step explanation:

Insert your x into the equation, then simplify for y.

5 0
3 years ago
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