1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
patriot [66]
3 years ago
15

Which statements are true about the trapezoid? not sure if i’m right

Mathematics
2 answers:
taurus [48]3 years ago
8 0
Yes you are absolutely right
Elena-2011 [213]3 years ago
6 0
From what I can see you are correct!
You might be interested in
Help, pls. In an algebra test rn
Nata [24]
I cannot see the picture
6 0
3 years ago
A baby weights 18 pounds at her four month appointment. Six months later she weights 24 pounds. By what percentage did the baby’
Anna11 [10]

Answer: 25%

Step-by-step explanation:

Given : Previous baby's weight = 18 pounds

New baby's weight =  24 pounds.

Increase in weight = New weight -Previews weight

=24 pounds - 18 pounds  =6 pounds

Percentage increase in baby's weight ==\dfrac{\text{Increase in baby's weight}}{\text{previous weight}}\times100

\\\\=\dfrac{6}{24}\times100=25\%

Hence, the percentage increase in baby's weight = 25%

4 0
3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

Z = \frac{X - \mu}{\sigma}

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
Children pick their favorite sports team 35% pick basketball 28% pick hockey what percent pick football
Kay [80]
37% pick football

35+28=63
100-63=37
7 0
2 years ago
I NEED HELP RIGHT NOW PLEASE
irina [24]

Answer:

35

Step-by-step explanation:

The entire angle is 70 degrees. If it is an angle Bi (which means two) sector, then divide it by two, which becomes 35 degrees.

8 0
3 years ago
Other questions:
  • How do you simplify the expression 7 - 2(2 + 5) -8?
    8·2 answers
  • 2( 4c - 7 ) < 8 ( c - 1) - 6 if possible
    12·1 answer
  • There are 16 peices of fruit in a bowl and 12 of them are apples. What percentage of the pieces of fruit in the bowl are apples?
    11·1 answer
  • consider the polynomial equation x(x-3)(x+6)(x-7)=0. which of the following are zeros of the equation? select all that apply.
    13·1 answer
  • Consider the following intermediate chemical equations.
    12·1 answer
  • There are 80 students total in 7th grade, of whom 15% are left handed. How many
    12·1 answer
  • Name the component that changes its resistance as the temperature changes.​
    6·2 answers
  • Raymond planted a vegetable garden containing 50 plants. There were 15 tomato
    13·1 answer
  • Please help if you can please
    12·1 answer
  • Round 3 047 398<br> to the nearest 1 000 000
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!