a trapezoid has an area of 50 square inches. The bases are 3 inches and 7 inches. What is the height of the trapezoid.

A chef turned on a faucet and filled a 2-gallon pot in 5 6 of a minute. What is the rate of water flow? Simplify your answer and

NARA [144]

Answer: Rate of water flow is 12/5 gallon per minute

Step-by-step explanation:

Given the following :

Volume of faucet pot = 2 gallon

Time taken to fill the faucet pot = 5/6 of a minute = (5/6) minute.

Rate of water flow:

Volume of faucet pot / time taken

= 2 gallon ÷ (5/6) minute

= 2 gallon × (6/5) minutes

= (12 /5) gallon / minute

= 2 2/5 gallon / minute

Hence rate of water flow is 12/5 gallon per minute.

7 0

3 years ago

Read 2 more answers

PLEASE HELP ME WITH PROOFS . DONT ANSWER IF YOU DONT KNOW. BRAINLY FOR CORRECT ANSWER

ZanzabumX [31]

Step-by-step explanation:

Does this help?

6 0

2 years ago

A music industry researcher wants to estimate, with a 90% confidence level, the proportion of young urban people (ages 21 to 35

katrin2010 [14]

Answer:

21378

Step-by-step explanation:

The confidence level is 90%, which gives us a Z of 1.645.

Our equation is (Z^2*p*q)/e^2

(1.645^2*p*(1-.21))/ (.01^2)

E is the error proportion, .01

Q is 1-p or the proportion

P is the pop you need

With this, we can solve

and get p = 21377.5975 and rounded to 21378

5 0

3 years ago

What is the measure of angle 1

creativ13 [48]

Answer:

86

Step-by-step explanation:

if u add the angle #'s on the INSIDE, itll give 94, since a triangle isnt obtuse, it would have to add up to 180, subtract 94 from 180 and u get the answer :)

hope this helps

8 0

3 years ago

Read 2 more answers

Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, percent of the

Arturiano [62]

Answer:

Step-by-step explanation:

From the given information:

We can compute the null hypothesis & the alternative hypothesis as:

${H_o}:\text{Distribution of snowfalls in her hometown is similar to claimed percentage }$

${H_a}:\text{Distribution of snowfalls in her hometown is not similar to claimed percentage }$

The degree of freedom = n - 1

The degree of freedom = 5 - 1

The degree of freedom = 4

At the level of significance of 0.05 and degree of freedom 4,

the rejection region = 9.488

However, we can compute the chi-square X² goodness of fit test as:

months frequency (p) observed O Expected E Chi-square $X^2= \dfrac{(O-E)^2}{E}$

Dec 0.2 16 16 $\dfrac{(16-16)^2}{16} =0$

Jan 0.250 11 20 $\dfrac{(11-20)^2}{20} =4.050$

Feb 0.200 16 16 $\dfrac{(16-16)^2}{16} =0$

Mar 0.200 18 16 $\dfrac{(18-16)^2}{16} =0.250$

Apr 0.150 19 12 $\dfrac{(19-12)^2}{12} =4.083$

Total 1.000 80 80 8.3833

∴

The test statistics X² = 8.3833

Thus; we fail to reject the $H_o$ since test statistics X² doesn't fall in the rejection region.

Therefore; there is sufficient evidence to conclude that the distribution of snowfalls in her hometown is not similar to the claimed percentage.

6 0

3 years ago