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3 years ago

Please help me!! It's some sort of PreAP math stuff

Mathematics

1 answer:

olga55 [171]3 years ago

3 0

Answer:

12x + 40y

Step-by-step explanation:

4(3x+10y)

multiply the outer number (4) with all the terms inside the parentheses

4 X 3x + 4 X 10y

12x + 40y

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Two planes, which are 3740 miles apart, fly towards eah other. Their speeds diifer by 35 miles per hour. If they pass eah other

solmaris [256]

Answer:

In 4 hours they flew a combined 3740/4 = 935mph or 467.5mph each. +/- 17.5 to this number,

485mph & 450mph are their speeds.

Step-by-step explanation:

I believe this is correct, if not feel free to let me know and I will fix it. I'm sorry in advance if it is incorrect.

4 0

3 years ago

Geometry help please! this is my last try! The two triangles below are similar. What is the similarity ratio of ΔABC to ΔDEF?

IRISSAK [1]

I will say C 2:1

if your left triangle is the first one, then it is 2

right triangle is 1

for the left triangle is twice as large as the right

6 0

3 years ago

Let E=IR be the formula for voltage in a circuit, where E = volts, I =

Varvara68 [4.7K]

Answer:

The answer is ohms=4

Step-by-step explanation:

substitute the letters with numbers and solve:

E=IR

12=I 3

12/3=4

3 0

3 years ago

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

erica [24]

Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$

$\Rightarrow P(F)=(1-0.9)+(1-0.8)+(1-0.7)=0.6$

Lef $F_2$ be the event that she failed the 2nd exam, so

$P(F_2)=1-\frac{P(E_2)}{P(E_1)}$

$\Rightarrow P(F_2)=1-0.8=0.2$

So, the conditional probability that she failed the 2nd exam is

$P\left(\frac{F_2}{F}\right)=\frac{P(F_2)}{P(F)}$

$\Rightarrow P\left(\frc{F_2}{F}\right)=\frac{0.2}{0.6}=\frac{1}{3}=0.33$

5 0

3 years ago

A pasture is twice as long as it is wide. its area is 88,200 ft2. how wide is the pasture?

OLga [1]

If the width is w and the length is l, then 2w=l. In addition, l*w=88200 (using the area formula). Plugging 2w in for l, we get 2w*w=88200=2w^2. Dividing both sides by 2, we get w^2=44100 and w=sqrt(44100)=210.

4 0

3 years ago