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pentagon [3]
3 years ago
8

1 Simplify the following expression. 2482 832 A. 23 8 2 OB. 2585 Oc. 8 23 OD. 25 85​

Mathematics
1 answer:
NikAS [45]3 years ago
7 0

Answer:

jhigufytduhguiyoutyfdtyxrfdgfhcjvhlkuiyftdyrtzfsgvhuyiftucgc

Step-by-step explanation:

jbhvcxzddgnxfh gnvb,

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There are (4^9)5 ⋅ 4^0 books at the library. What is the total number of books at the library?
LenKa [72]
Your answer is B. You multiply 5 by the exponents and 4^0 is just 1, so 4^45•1=4^45. Hope this helped!
7 0
3 years ago
The price of a theater tickets 56.89.There are 22 seats in each row of the theater .How much would it cost to buy tickets for 2
mihalych1998 [28]

Answer:

Step-by-step explanation:

one row has 22 seats

2 rows have 44 seats

44x56.89=2,503.16

5 0
2 years ago
Read 2 more answers
What is the value of x?
Alexus [3.1K]

Answer:

x = 22.5

Step-by-step explanation:

\frac{BH}{DH}  =  \frac{HM}{HT}  \\  \frac{6}{15 + 6}  =  \frac{9}{9 + x}  \\  \frac{6}{21}  =  \frac{9}{9 + x}  \\ 6(9 + x) = 9 \times 21 \\ 54 + 6x = 189 \\ 6x = 189 - 54 \\  6x = 135 \\ x =  \frac{135}{6}  \\ x = 22.5

5 0
3 years ago
A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a stand
Art [367]

Answer:

b. Mean = 1.6 years, standard deviation - 0.92 years, shape: approximately Normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal Variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a standard deviation of 4.2 years. Sample of 35:

This means that:

\mu_G = 15

s_G = \frac{4.2}{\sqrt{35}} = 0.71

The distribution of life spans for electric ranges has a mean of 13.4 years and a standard deviation of 3.7 years. Sample of 40:

This means that:

\mu_E = 13.4

s_E = \frac{3.7}{\sqrt{40}} = 0.585

Which of the following best describes the sampling distribution of the difference in mean life span of gas ranges and electric ranges?

Shape is approximately normal.

Mean:

\mu = \mu_G - \mu_E = 15 - 13.4 = 1.6

Standard deviation:

s = \sqrt{s_G^2+s_E^2} = \sqrt{0.71^2+0.585^2} = 0.92

So the correct answer is given by option b.

8 0
3 years ago
There were 55 questions on your chemistry exam , your score was a 94.5 %. how many questions did
pav-90 [236]

Answer:

You got about 52 (~51.975) questions right on your chemistry exam

You got a 93.1% on your biology exam

Step-by-step explanation:

55 multiplied by 0.945 gives you about 52

And for the second part just divide 54/58 to get the percentage

6 0
2 years ago
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