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3 years ago

Please help me solve this asap. (step by step explanation)

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

5 0

First off, we see the y-intercept is at (0, 240), allowing us to eliminate F and H.

Then we look for the slope: the line decreases from 240 to 210 (a change of -30) when traveling 300 miles (a change of 300). Since the slope is rise over run, we have -30/300=-1/10.

Therefore, the correct answer is J.

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HELP ASAP!!!!!!!!!!!!

torisob [31]

Answer:

163 in²

Step-by-step explanation:

To find the area, I divided the shape into squares and rectangles. Then, I found the area of the squares and rectangles and added them,

Area = (7 x 7) + (18 x 5) + (6 x 4)
=> (49) + (90) + (24)
=> 163 in²

Hence, the area is 163 in.²

7 0

2 years ago

Will give brainliest!!

yawa3891 [41]

Answer:

$r=\frac{9\cos \theta}{\sin^2 \theta}$

Step-by-step explanation:

The given equation in rectangular coordinates is;

$y^2=9x$

We use the relation;

$x=r\cos \theta$

and

$y=r \sin \theta$

This implies that;

$(r \sin \theta)^2=9(r \cos \theta)$

$r^2 \sin^2 \theta=9r \cos \theta$

Divide through by r to get;

$r \sin^2 \theta=9\cos \theta$

Divide both sides by $\sin^2 \theta$

$r=\frac{9\cos \theta}{\sin^2 \theta}$

8 0

4 years ago

Victor draws one side of equilateral ∆PQR on the coordinate plane at points P (–9, –2) and Q (–2, –2). What are the two possible

Elenna [48]

Answer:

the vertices of point R could be (-2,5) or (-2,9)

each point is 7 units away from Q since there are 7 units between P and Q

Step-by-step explanation:

6 0

3 years ago

A train left a station a station at noon going 40mph. A second train left at 3 p.m. going 60mph. At what time will the second tr

Sati [7]

Here speed of first train is slower than speed of second train. So if they catch up , the distance traveled will be same since they left the same station at different times.

But the time traveled will be different .

Let us say the two trains meet at time t ,

Distance = speed * time

Distance traveled by first train = 40 t

Since second train is leaving after three hours , so

Distance traveled by second train = 60(t-3)

As discussed earlier, distance traveled will be same, so

$40t =60(t-3)$

distributing 60 over t-3

$40t =60t-180$

$60t -40t=180$

$20t =180$

$t =180/20$

t=9

So the second train will meet the first at 9 p.m.

5 0

3 years ago

Read 2 more answers

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300. The

zubka84 [21]

Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

$P(X = 0) =1 * 1* (1-0.2)^{7$

$P(X = 0) =(0.8)^{7$

$P(X = 0) =0.2097152$

If it rains on any one day; then X = 100

$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

$P(X = 100) = _7C_1 * 0.2^1 * (1-0.2)^{7-1$

$P(X = 0) =7 * 0.2* (1-0.2)^{6$

$P(X = 100) =7* 0.2* (0.8)^{6$

$P(X = 100) =0.3670016$

if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

$P(X = 200) = 21 * 0.2^2 * (0.8)^{5$

$P(X = 200) = 0.2752512$

if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

$(E [X]) = \sum x * p\\ \\ (E [X]) = 0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55$

NOW;

the variance of the refund payment to the couple = 28002.944 - 18539.55

the variance of the refund payment to the couple = 9463.394

7 0

3 years ago