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Murljashka [212]
3 years ago
14

Help i don't know what i'm doing​

Mathematics
2 answers:
yarga [219]3 years ago
7 0

Answer: don't worry always pray for help

Step-by-step explanation:

sashaice [31]3 years ago
7 0

Answer:

C

Step-by-step explanation:

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PLZ HELP ASAP AREA OF A SECTOR
const2013 [10]
The fraction of a circle represented by the shaded region is:
 S '/ S = ((8/5) * pi * r) / (2 * pi * r)
 Rewriting we have:
 S '/ S = ((8/5)) / (2)
 S '/ S = 8/10
 S '/ S = 4/5
 Then, we can make the following rule of three:
 (64/5) pi -------> 4/5
 x ----------------> 1
 Clearing x we have:
 x = ((1) / (4/5)) * ((64/5) pi)
 Rewriting:
 x = (5/4) * ((64/5) pi)
 x = (5/4) * ((64/5) pi)
 x = 50.24
 Answer:
 
The area of the complete circle is:
 
x = 50.24
4 0
3 years ago
Find the final amount of money in an account if
LuckyWell [14K]

Answer:

10213.33

Step-by-step explanation:

use the formula to answer this question

principal=7,200

rate=0.06

compounded anually=1

time=6

Hope it helped..

3 0
3 years ago
You have a small business making and delivering box lunches. You calculate your average weekly cost y of producing x lunches usi
kogti [31]

250 lunches are produced by the small business in last week.

<u>Step-by-step explanation:</u>

It is given that,

  • y ⇒ the average cost per week.
  • x ⇒ the number of lunches produced per week.

The function relating these two factors x and y is given as y = 2.1x + 75

  • The cost of the last week is y = $600.
  • The lunches made last week is x = unknown.

<u>To find the value of x :</u>

Substitute y= 600 in the given function,

⇒ 600 = 2.1x + 75

⇒ 2.1x = 600 - 75

⇒ x = 525 / 2.1

⇒ x = 250

Therefore, the lunches prepared last week is 250.

6 0
4 years ago
Find the sum of the given polynomials.
enot [183]
Number 4 is your answer
8 0
4 years ago
Read 2 more answers
Weights were recorded for all nurses at a particular hospital, the mean weight for an individual nurse was 135 lbs. with a stand
Aloiza [94]
Given:
population mean, &mu; =135
population standard deviation, &sigma; = 15
sample size, n = 19

Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.

Estimate of sample mean
\bar{x}=\mu=135

Estimate of sample standard deviation
\s=\sqrt{\frac{\sigma^2}{n}}
=\sqrt{\frac{15^2}{19}}=3.44124  to 5 decimal places.

Thus, using the normal probability table,
P(125
=P(\frac{125-135}{3.44124}
=P(-2.90593
=P(Z
=P(Z

Therefore 
The probability that the mean weight is between 125 and 130 lbs 
P(125<X<130)=0.0731166-0.0018308
=0.0712858



3 0
3 years ago
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