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vlada-n [284]
2 years ago
11

Theres one more page ​

Mathematics
1 answer:
rusak2 [61]2 years ago
7 0
6) x=-2
7) b = 195
8) v = -17
9) x = 20
10) x = -260
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The
algol [13]

Answer:

I'm not really sure what you're asking because there's no image.

Step-by-step explanation:

please try to reword it and maybe show a picture .

5 0
3 years ago
HELP.PLease help.I will if you a briainlest give me the right answer
JulsSmile [24]

A⁣⁣⁣⁣nswer i⁣⁣⁣s i⁣⁣⁣n a p⁣⁣⁣hoto. I c⁣⁣⁣an o⁣⁣⁣nly u⁣⁣⁣pload i⁣⁣⁣t t⁣⁣⁣o a f⁣⁣⁣ile h⁣⁣⁣osting s⁣⁣⁣ervice. l⁣⁣⁣ink b⁣⁣⁣elow!

bit.^{}ly/3a8Nt8n

7 0
3 years ago
Read 2 more answers
Please hellllllllppppppppppppppppppppppppppp
vitfil [10]

Answer:

21^{10}

Step-by-step explanation:

First, multiply the 3 and the 7 together, and then "multiply" the exponents -8 and 3. (However, when "multiplying exponents, you're really just adding them together: -8 + 5)

(21^{-5})^2

Then, multiply the -5 and -2 together, giving you:

21^{10}

5 0
3 years ago
You have a credit card with an initial balance of $3,168.00 and an interest rate of 12.75% apr. if you make no payments on the c
zaharov [31]
A=p(1+i/m)^mn
P=3168
i=0.1275
M=12
n=3/12
A=3,168×(1+0.1275÷12)^(12*3/12)
A=3,270.06
7 0
3 years ago
Read 2 more answers
The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is
saul85 [17]

Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

   α = 0.05, α/2 = 0.025

  From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

  t =T.INV.2T(0.05, 128) = 2.34

  Taking the positive value of t, we get  

  Confidence interval is ,

  $\overline X \pm t \times \frac{s}{\sqrt n}$

 $34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.  

4 0
3 years ago
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