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3 years ago

Jessie pays $2 for each carnival ride plus $8

1 answer:

3 0

$ 2 each carnival ride + $8 spend at most $35

Rewrite that is
$2x + 8 ≤ 35

I hope this helped and that I was right, have a nice day.

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Use the circular mirror shown

aliina [53]

Multiply 15 x 3.14. You'll get 47.1. Thats the area

3 0

2 years ago

Can someone help with number 20

enyata [817]

Answer:B: 36°

Step-by-step explanation:

We know that ∆ABC is isoceles, making (angle)<ABC and <BCA congruent because base angles of isoceles triangles are congruent.

Because we have parallel lines, we can look for alternate interior angle pairs. <BCA is congruent to <DAC because they're alternate interior angles.

If <BCA is x then so is <ABC.

Since triangles add up to 180° we can add all of the angles (3x+x+x) and set it equal to 180.

3x+x+x=180

5x=180

x=36

If we were looking for <BAC we would plug that back in and solve, but we're looking for <BCA which is equal to x, therefore m<BCA=36°

3 0

3 years ago

Round 705,499 to the nearest thousand.

makvit [3.9K]

Answer:

705,000

Step-by-step explanation:

To round to the nearest thousand, you look at the number to the right of 5. If the number is 5 or higher then you round 5 up. But it is 4 so you keep 5 the same. All of the numbers after 5 are now 0. 705,000

3 0

2 years ago

Read 2 more answers

Simplify the following expression (x^5)^2=x^n A. 7 B. 10 C. -2 D. 9

Evgen [1.6K]

Answer:

This is the farthest I could go in solving the equation. Is there anymore information about the problem? Unless you are looking for n which it would seem n would equal 10.

4 0

3 years ago

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.

8 0

3 years ago