8. A length of rope measures 5/8 of a yard long. Rylan would like to cut it into two equal portions.

A certain forest covers an area of 2400 km^2. Suppose that each year this area decreases by 5.75%. What will the area be after 8

kolezko [41]

We have to use the function A(t)= a(1 +/- r)^t
So A is going to be the answer- the area after 8 years. t is going to be 8, because t is time. a is 2400 because that's the starting value. Finally, we're going to turn 5.75% into it's decimal form of .0575. That's going to be r, and because the area is decreasing, we will subtract r from 1.
Our new function is:
A=2400(1-.0575)^8
Simplify a little bit.
A=2400(.9425)^8
Bring .9425 to to power of 8 and then multiply by 2400 because PEDMAS.
A=1494.38366866
The rounded answer is 1494 kilometers squared.

5 0

3 years ago

Which number line shows the correct placement of ‐200‐200, 7575, and ‐150‐150?

Anna35 [415]

What are the choices here

7 0

3 years ago

At the arcade, nick spent 203 tokens on videogames in seven days. each day he spent five more tokens than the previous day. how

Dafna11 [192]

Suppose nick spent x number of tokens on day 1

So he spent x + 5 on day 2

He spent x + 10 on day 3

He spent x + 15 on day 4

He spent x + 20 on day 5

On day 6 he spent x + 25

And on day 7 he spent x + 30

So in all he spent

x + x + 5 + x + 10 + x + 15 + x + 20 + x + 25 + x + 30

= 7x + 105

But total token spent = 203

So

7x + 105 = 203

Subtracting 105 from both sides

7x = 203 - 105

7x = 98

Dividing by 7 on both sides

x = 14

So on third day he spent x + 10

= 14 + 10

= 24 tokens

Tokens spent on day 3 = 24

8 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

I am a multiple of 7 and divisible by 4 if i am less than 30 what number am i?

SOVA2 [1]

Answer: 28

Step-by-step explanation:

The multiples of 7 that's less than 30 include: 7, 14 , 21 and 28.

From the numbers above,

7 isn't divisible by 4 as it gives a remainder.

14 isn't divisible by 4 as it gives a remainder.

21 isn't divisible by 4 as it gives a remainder.

28 is divisible by 4 without a remainder

Therefore, the correct answer should be 28

3 0

2 years ago