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3 years ago

Help please with my geometry

Mathematics

1 answer:

Alja [10]3 years ago

6 0

X is equal to 4

Because RS + ST=RT
So basically x+3x=5x-4
4x=5x-4
-4x -4x
-1x=-4
x=4

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The Jones family is landscaping their yard. Their yard yard is 160 square feet and one side is 10 feet long. What is the length

Stels [109]

The length of the other side of the yard is 16 feet
work: 160 divided by 10
hope it helps

3 0

3 years ago

What do you add to 3 4/7 to make 6??

Alla [95]

You'll need to add 2 3/7 to <span> 3 4/7 to make 6.
3 + 2 = 5 4/7 + 3/7 = 7/7 =1
5+1=6</span>

6 0

4 years ago

A company has an opportunity to bid on three contracts. Determine which would be the best investment given the information in th

MariettaO [177]

CONTRACT PROFIT, (P)* OF PROFIT...(P)*. TO BREAK EVEN...LOSS, (P)* OF LOSS

Southeast $45,000, 50% 30% $6,000, 20%

Southwest $60,000, 35% 40% $10,000, 25%

California $112,000, 20% 40% $40,000, 40%

I think that the best contract would be with SOUTHEAST. It probability of having a profit is higher than the other two at 50% while its probability of incurring a loss is lesser than the other two at 20%.

Despite the bigger value of its contract profit of the other two, these value will still be greatly affected by its corresponding probabilities.

6 0

3 years ago

Read 2 more answers

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

Free 50 point your welcome fam

Travka [436]

Answer:

That pretty cute

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers