If e and d are midpoints of triangle abc find cb

The weights of cars passing over a bridge have a mean of 3,550 pounds and

Dovator [93]

The approximate probability that the weight of a randomly-selected car passing over the bridge is more than 4,000 pounds is 69%

Option C is the correct answer.

<h3>What is Probability ?</h3>

Probability is defined as the study of likeliness of an event to happen.

It has a range of 0 to 1.

It is given in the question that

The weights of cars passing over a bridge have a mean of 3,550 pounds and standard deviation of 870 pounds.

mean = 3550

standard deviation, = 870

Observed value, X = 4000

Z = (X-mean)/standard deviation = (4000-3550)/870 = 0.517

Probability of weight above 4000 lb

= P(X>4000) = P(z>Z) = P(z> 0.517) = 0.6985

The approximate probability that the weight of a randomly-selected car passing over the bridge is more than 4,000 pounds is 69%

To know more about Probability

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6 0

1 year ago

A number pattern starts with 10 and follows the rule multiply by 3 What is true

lesya692 [45]

Answer:

B) They have a 0 in the ones place.

Step-by-step explanation:

Given

Number pattern that starts with 10.

Rule: Multiply by 3

Required

Select which statement is truth of the pattern.

First, the pattern needs to be listed out.

Given that the pattern starts from 10 and progresses by multiplying current term by 3.

The pattern is as follows;

10, 30, 90, 270, 810, 2430, 7290, .....

By comparing the given options with the above pattern of numbers. We have:

A) They are odd numbers.

This statement is false because the above pattern of numbers are all even and can never be odd

B) They have a 0 in the ones place.

This statement is truth because all the numbers in the above progression ends in 0. This 0 represents the unit of one's place.

Hence, this statement is truth.

C) They have a 3 in the tens place.

This statement is false;

This is so because they have different digits at tens place. The tens place is the digit prior to the last digit of each number.

D) They can be odd or ever numbers.

As pointed out in (A) above, the number can only be even number because the progression starts with an even number (10) and irrespective of what it's been multiplied by, it'll always remain an even number.

Hence,

Only option B is correct/truth.

Others are incorrect

3 0

3 years ago

How many pounds does 6.5 dozen cupcakes weigh?<br> • 1 cupcake = 35.3 grams

melisa1 [442]

Answer:

35.3g * 78 =2,753.4g

2,753.4g = 6.07lbs

6 0

2 years ago

Work out<br> (9×10²)+(7×10³)<br> State your answer in standard form

Sunny_sXe [5.5K]

Answer:

(9×10×10)+(7×10×10)

900+700

1600

1.6×10^3

4 0

2 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

2 years ago