All categories

3 years ago

Help please..............

Mathematics

1 answer:

eimsori [14]3 years ago

6 0

I believe the correct answer here would be B. The values x=2 and y=5 make the equation true. The answer is not A, because the graph could also go through 3 and 6. It is not C because that is not the correct order. The answer is not D because D is absurd and does not make sense.

You might be interested in

Mandy has an annual salary of $37,580. Each month she has a car payment of $265 and a student loan of $120. If she applies for a

lukranit [14]

Answer: Not likely; recurring debt is higher than what is allowed.

Step-by-step explanation:

The options are:

a. Very likely; recurring debt is less than what is allowed.

b. Somewhat likely; recurring debt is equal to what is allowed.

c. Not likely; recurring debt is higher than what is allowed.

d. There is not enough information given to determine the answer.

Annual salary = $37,580

Car loan payment annually = $265 × 12 = $3180

Annual student loan= $120 × 12 = $1440

Annual loan payment will be:

= $3180 + $1440

= $4620

Based on the above, Mandy is already paying 12.3% (4620 / 37580)× 100 as loan every month. Coupled with the fact that he'll still have some amount to pay as loan every month, he won't be given as the recurring loan is higher than what will be allowed.

5 0

3 years ago

PLEASE HELP I AM BEHIND!!!!

romanna [79]

Answer:

The answer is c definitely

6 0

3 years ago

Read 2 more answers

Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minute

Margaret [11]

Answer:

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

The problem states that:

Mean = 35, so $\mu = 35$

Variance = 81. The standard deviation is the square root of the variance, so $\sigma = \sqrt{81} = 9$ .

Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.

Looking at the z-score table, $z = -0.44$ has a pvalue of 0.333. So what is the value of X when $z = -0.44$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{X - 35}{9}$

$X - 35 = -3.96$

$X = 31.04$

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of $X = 40.5$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 35}{9}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if $P_{1} = 100% - 72.91% = 27.09% = 0.2709$

$P_{2}$ is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of

$X = 38.5$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38.5 - 35}{9}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if $P_{1} = 100% - 65.17% = 34.83% = 0.3483$

So:

$P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192$

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

7 0

4 years ago

A cargo ship traveled 150 nautical miles in 6 hours at a constant speed. How far did the cargo ship travel in one hour

mixas84 [53]

Answer:

25 nautical miles

Step-by-step explanation:

We need to find the rate. We can do so using the formula, distance = rate * time

d= rt

We know the distance and the time, so we need to solve for the rate

d/t = rate

d= 150 nautical miles

t = 6 hours

d/t = 150 nautical miles /6 hours

rate = 25 nautical miles per hour

So in 1 hour we can go 25 nautical miles

7 0

3 years ago

Read 2 more answers

عی<br> --<br> What is 6/7•1/3

denis23 [38]

Answer:

14/41=9

Step-by-step explanation:

6*3/7*3=18/21

1*7/3*7=7/21

18*7=126/441

126/9=14

441/9=49

7 0

3 years ago