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ale4655 [162]
3 years ago
10

It math bad at it please help:);-)

Mathematics
1 answer:
abruzzese [7]3 years ago
6 0

Answer:

It is -1/7 x (6p-1)

Step-by-step explanation:

Btw download photomath on your phone it helps alot

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Distance between home and school(on map)=5cm

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therefore distance between home and school=
5/4×8 = 1.25×8 = 10km
5 0
4 years ago
If x²+3x=5 what is the value of x?​
lions [1.4K]

Answer:

Its C trust

Step-by-step explanation:

7 0
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.

8 0
4 years ago
Use a calculator to solve for xin the equation 2e^3x = 400. Round your answer to<br> three decimals.
Paul [167]

the answer is

D. 1.766
3 0
4 years ago
Find the area of the trapezoid.<br> 40 mm<br> 24 mm<br> 19 mm<br> A=<br> mm2
UNO [17]

Answer:

A = 708 mm^2

Step-by-step explanation:

A=\frac{a+b}{2} h\\\\A=\frac{19+40}{2} *24\\\\A=\frac{59}{2} *24\\\\A= 708

5 0
3 years ago
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