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2 years ago

Can someone please help?

Mathematics

1 answer:

maria [59]2 years ago

3 0

x = 3x - 70 as corresponding angles

2x = 70

x = 35

∠LMN = 3x - 70 = 3*35 - 70 =35⁰

∠LMN = 35⁰

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Which fraction is greater 7 out of 11 or eight out of 11

Ahat [919]

Answer:

8 out of 11

Step-by-step explanation:

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2 years ago

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Cómo simplificar 2/3 paso a paso<br>

slamgirl [31]

Answer:

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2 years ago

Which graph shows the solution set for Negative 1.1 x + 6.4 greater-than negative 1.3? A number line going from negative 10 to 0

Rufina [12.5K]

Answer:

A number line going from 0 to 10. An open circle is at 7. Everything to the left of the circle is shaded.

Step-by-step explanation:

We have the following inequality:

-1.1x + 6.4 > -1.3

to solve it, we need to isolate x, as follows:

-1.1x > -1.3 - 6.4

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3 years ago

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Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha }{2} } = 1.96$

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

Then the critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

$\= x - E < \mu < \= x + E$

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

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2 years ago

An engineering school reports that 52% of its students are male (M), 33% of its students are between the ages of 18 and 20 (A),

weeeeeb [17]

Answer:

42%

Step-by-step explanation:

Given: P(M) = 0.52, P(A) = 0.33, and P(M and A) = 0.27.

Find: P(not M and not A).

P(not M and not A) = 1 − P(M or A)

P(not M and not A) = 1 − (P(M) + P(A) − P(M and A))

P(not M and not A) = 1 − (0.52 + 0.33 − 0.27)

P(not M and not A) = 1 − 0.58

P(not M and not A) = 0.42

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3 years ago