Please help me answer this!!!

the grocery store is 3.76 kilometers from ryan house and 1.67 kilometers from abby house how much farther does ryan live from th

elena55 [62]

The question is basically asking for the difference between how far ryan lives from the store and how far abby lives from the store.

therefore, to solve the problem, we can just do: distance ryan lives from the store minus distance abby lives from the store.

since ryan is 3.76 kilometers from the store and abby is 1.67 kilometers from the store, the answer is:

3.76 - 1.67 = 2.09 kilometers.

4 0

3 years ago

Angel is 35 years old. She has $550 in a checking account, $6,500 in investments, $15,000 in a retirement account, and owns a ho

krek1111 [17]

$7,050. Because the term "liquid Assets" refer to money, investments, and generally things that can be liquidated into direct cash in less than 30 days. So if you add the $550 in the Checking Account, and $6500 in investments. You get $7050

7 0

3 years ago

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

The two triangles are similar. What is the value of x? Enter your answer in the box.

ra1l [238]

Just took the test and got the the correct answer *\(♡°▽°♡)/*

Look at the image down below!!

5 0

3 years ago

Read 2 more answers

At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist headi

gizmo_the_mogwai [7]

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Find the duration
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Time : 9:00am to 10:15am
Duration: 1hour 15 mins or 1.25hours

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Find difference in distance travelled between the two cyclists.
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Difference in speed = 6km/h
1 hour = 6 km
1.25 hours = 7.5km

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Find the distance the cyclist traveled
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42.5 - 7.5 = 35km
35km ÷ 2 = 17.5km

The slower cyclist traveled 17.5 km in the 1.25 hours

17.5 + 7.5 = 25 km

The faster cyclist traveled 25km in the 1.25 hours

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Find the speed of the slower cyclist
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1.25 hour = 17.5 km
1 hour = 17.25 ÷ 1.25
1 hour = 14km

The cyclist was traveling at 14km/h

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Find the speed of the faster cyclist
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1.25 hour = 25km
1 hour = 25 ÷ 1.25
1hour = 20 km

The cyclist was traveling at 20km/hour

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Answer: The speed were 14km/h and 20 km/h
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8 0

3 years ago

Read 2 more answers