On a map, 2 inches represents 300 miles. How many inches would there be between two cities that are 1,200 miles apart?

How ? rn i’m doing a warm up

levacccp [35]

Answer:

For question y=x+3

Your answer is y=(2x+1)+3

Step-by-step explanation:

5 0

3 years ago

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

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of sc

mr Goodwill [35]

<span>In a normal distribution 68.27% of the values are within one standard deviation from the mean, 95.5% of the values are within two standard deviations from the mean, and 99.7 % of the values are within three standard deviations of the mean

With that you have the answer to the three questions:

</span>
<span>a. significantly high (or at least 2 standard deviations above the mean).

99.5% of the values are within 2 standard deviations from the mean, half of 100% - 95.5% = 4.5% / 2 = 2.25% are above the mean, so the answer is 2.25%

b. significantly low (or at least 2 standard deviations below the mean).

The other half are below 2 standard deviations, so the answer is 2.25%

c. not significant (or less than 2 standard deviations away from the mean).

As said, 95.5% are within the band of two standard deviations from the mean, so the answer is 95.5%.
</span>

7 0

3 years ago

Read 2 more answers

The students in the Service Club are mixing paint to make a ural. The table below

maw [93]

Answer:

two different shades of what I think is pink.

1/3 is equal to 2/6 which is equal to 3/9

1/2 is equal to 2/4 which is equal to 4/8

The bolded ones are the original mixtures given.

Start with 1/3 Multiply top and bottom by 2 and you get 2/6

When you multiply top and bottom of 1/3 by by 3 you get 3/9

Use the same idea for 1/2

Step-by-step explanation:

Start with 1/3 Multiply top and bottom by 2 and you get 2/6

When you multiply top and bottom of 1/3 by by 3 you get 3/9

Use the same idea for 1/2.

4 0

3 years ago

Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are su

posledela

Answer: a) Independent

b) Independent

c) Dependent

Step-by-step explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a) $E_1$ : tails comes up with the coin is tossed the first time;

$E_1$ = { TTT, THH, THT, TTH }

$E_2$ : heads comes up when the coin is tossed the second time.

$E_2$ = { THT, HHH, THH, HHT }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{2}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Since, $E_1\cap E_2$ = { THH, THT }

$n(E_1\cap E_2) = 2$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}$

Thus, $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are independent events.

B) $E_1$ : the first coin comes up tails

$E_1$ = { TTT, THH, THT, TTH }

$E_2$ : two, and not three, heads come up in a row

$E_2$ = { HHT, THH }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{4}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since, $E_1\cap E_2$ = { THH }

$n(E_1\cap E_2) = 1$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}$

Thus, $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are independent events.

C) $E_1$ : the second coin comes up tails;

$E_1$ = { HTH, HTT, TTT, TTH }

$E_2$ : two, and not three, heads come up in a row

$E_2$ = { HHT, THH }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{4}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since, $E_1\cap E_2$ = $\phi$

$n(E_1\cap E_2) = 0$

⇒ $P(E_1\cap E_2) = 0$

Thus, $P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are dependent events.

3 0

3 years ago