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3 years ago

Can someone please help me with these math questions? I don’t get them.

Mathematics

1 answer:

Dovator [93]3 years ago

8 0

Answer:

X=108 degrees

Step-by-step explanation:

well first a triangle is made up of 180 degrees, therefore add up 42+30=72, then subtract 180-72=108 sorry dont know the second one

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It costs $2.50 to rent bowling shoes. Each game costs $2.25. You have $9.25. Write and solve an equation to find how many games

Zinaida [17]

The equation is 2.25x2.50.
You just have to put 9.25=2.25 x 2.50. Hope this helps!♡

7 0

4 years ago

Read 2 more answers

Carroll Corporation has two products, Q and P. During June, the company's net operating income was $22,500, and the common fixed

Alexus [3.1K]

Answer:

The segment margin for Product P = $41,000

Step-by-step explanation:

Total Segment Margin = Net Operating Income + common fixed expenses

= $ 25,000 + $ 37,000

= $ 62,000

Total Segment Margin = Segment Margin of Q + Segment Margin of P

$ 62,000 = $ 21,000 + Segment Margin of P

or Segment Margin of P = $ 62,000 - $ 21,000

= $ 41,000

5 0

4 years ago

I think it can be a triangle but want another opinion

pashok25 [27]

Answer:

I think It's a triangle too

Step-by-step explanation:

3 0

3 years ago

Given f (x) = 1/3 (x +6) what is f (-6)? help me pretty please!!

Blizzard [7]

Answer:

Step-by-step explanation:

Put -6 where x is and do the arithmetic.

f(-6) = 1/3 (-6 +6) = 1/3 (0)

f(-6) = 0

3 0

3 years ago

Read 2 more answers

A survey reports that 67% of college students prefer to drink more coffee during the exams week. If we randomly select 80 colleg

Akimi4 [234]

Answer:

The probability that at most 50 say that they drink coffee during exam week is 0.166.

Step-by-step explanation:

The random variable X can be defined as the number of college students who prefer to drink more coffee during the exams week.

The probability of the random variable X is p = 0.67.

A random sample of n = 80 college students are selected.

The response of every students is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 80 and p = 0.67.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=80\times 0.67=53.6>10\\n(1-p)=80\times (1-0.67)=26.4>10$

Thus, a Normal approximation to binomial can be applied.

$X\sim N(np, np(1-p))$

The mean of the distribution of X is:

$\mu=np=80\times 0.67=53.6$

The standard deviation of the distribution of X is:

$\sigma=\sqrt{np(1-p)}=\sqrt{80\times 0.67\times (1-0.67)}=4.206$

A Normal distribution is a continuous distribution. So, the probability at a point cannot be computed for the Normal distribution. To compute the probability at a point we need to apply the continuity correction.

Compute the probability that at most 50 say that they drink coffee during exam week as follows:

Apply continuity correction:

$P(X\leq 50)=P(X$

$=P(X$

*Use a z-table for the probability.

Thus, the probability that at most 50 say that they drink coffee during exam week is 0.166.

5 0

3 years ago