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1 year ago

6/7*(11/18-5/12) pls help: fraction form, please

Mathematics

1 answer:

olasank [31]1 year ago

7 0

Step-by-step explanation:

$\frac{6}{7} \times ( \frac{11}{18} - \frac{5}{12} )$

$\frac{6}{7} \times ( \frac{7}{36} )$

$\frac{42}{252}$

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Simplify expression 8x-6+x-1

charle [14.2K]

Answer:

9x -7

Step-by-step explanation:

8x-6+x-1

Combine like terms

8x +x -6-1

9x -7

8 0

3 years ago

Solve /x-5/=3 A. x=2, x=8 B. x=-8, x=8 C. x=-2, x=2 D. x=-8, x=-2

zhenek [66]

Answer:

A) x=2, 8

Step-by-step explanation:

abs(x-5)=3

x-5=3, x-5=-3

x=3+5=8,

x=-3+5=2

5 0

2 years ago

Germany has a population

egoroff_w [7]

Answer:

C. -186,205

Step-by-step explanation:

636,854 - 827,155 + 684,862 - 680,766

=> 1,321,716 - 1,507,921

=> -186,205

8 0

2 years ago

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$

Here

$\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(Z < 2.37 ) - P(Z < -2.37 )$

From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is

$P(Z < 2.37 ) = 0.9911$

and

$P(Z < - 2.37 ) = 0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) =0.9911-0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = 0.9822$

=> $P(| \^ p - p| < 0.05 ) = 0.9822$

3 0

3 years ago

#1. Which is equal to z3 8? A. (z - 2)(z^2 - 2z - 4). B. (z + 2)(z^2 - 2z + 4) C.(z - 2)(z^2+ 2z + 4). D. (z + 2)(z^2+ 2z - 4) #

tankabanditka [31]

#1. B
(z * z^2 + z * 2z + z * 4) – (-2 *z^2 – (-2) 2z – (-2) 4)
Z^3 + 2z^2 + 4z – 2z^2 -4z – 8
Z^3 + 2z^2 – 2z^2 + 4z – 4z – 8
Z^3 - 8

#2 and #3. D
(x + y)(x + 2)
x^2 + 2x + yx + 2y

#4. D.
(x - 7)(x + 7)(x- 2)
x^2 + 7x – 7x -49
x^2 + x – 49
x^2 -49
(x^2 – 49 ) (x – 2)
x^3 – 2x^2 – 49x + 98

#5. C
(y - 4) = 0
y = 4
(x + 3)= 0
x = -3

#6. A and B

3 0

3 years ago