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4 years ago

PLS HURRY!! Simplify. 7/3+3(2/3−1/3)^2 Enter your answer in the box.

Mathematics

2 answers:

Irina-Kira [14]4 years ago

8 0

Use order of operations (PEMDAS):-

7/3+3(2/3−1/3)^2 Deal with parentheses first:-

= 7/3 + 3 * 1/3^2 now exponent:

= 7/3 + 3 * 1/9 now the multiplication:-

= 7/3 + 3/9

= 7/3 + 1/3

= 8/3 Answer

AveGali [126]4 years ago

3 0

Do distributed property 7/3+3(2/3-1/3)^2 7/3+3×(2/3-1/3)^2 7/3+3×(1/3)^2 7/3×3×1/9 7/3×1/3 7+1/3 8/3 or 2 and 2/3 or 2.66667

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Answer:

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3 years ago

1/5 of the students at Kendall Middle School are in 6th grade. 3/7 of the students at Kendall Middle School are in 8th grade. Wh

ELEN [110]

13/35 because you have to find the common denominator which would be 35 and in order to get 7 to 35 you have to multiply the whole thing by 5 making the fraction equal 15/35 and you have to multiply the other fraction by 7 making it 7/35 add the two together which would equal 22/35 so to make that 35/35 it would have to be 13/35

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3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

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4 years ago

Which of the following is not a type of angle <br> a acute <br> b obtuse<br> c parallel<br> d right

scZoUnD [109]

Answer:

parallel

Step-by-step explanation:

parallel describes lines

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3 years ago

Read 2 more answers

If The quotient of three times a number and 4 is no more than 13.

alexandr1967 [171]

Answer:

3x + 4 ≤ 13

Step-by-step explanation:

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4 years ago