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neonofarm [45]
3 years ago
13

Point A is located at (0, 4), and point C is located at (−3, 5). Find the x value for point B that is located one over four the

distance from point A to point C. (4 points)
−0.25

−0.5

−0.75

−1
Mathematics
2 answers:
uysha [10]3 years ago
7 0
The answer is 0.25 hope this helps
Alex787 [66]3 years ago
3 0

Hi!

Answer:

C.-0.75

Step-by-step explanation:

A(0,4) C(-3,5)

distance between x-coordinates is 3

distance between y-coordinates is 1

We're asked to find point B which

is located 1/4 distance from point A to point C.

So we will take the distance of the x-coordinates and y-coordinates and multiply them by 1/4.

3 x 1/4 = 0.75

1 x 1/4 = 0.25

Now, we are looking for the x value in which point B is located, on a coordinate grid C is to the left of A, which means this is a negative movement. Thus, all we have to do is subtract 0.75 from 0 since it is negative.

0 - 0.75 = -0.75

Then that's

the x value for point B.

Hope this helped!

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Can somebody pls help I’ll mark brainliest
Mnenie [13.5K]
<span><u><em>Answer:</em></u>
64

<u><em>Explanation:</em></u>
The square of any number can be obtained by multiplied the number by itself.
<u>In other words:</u>
square of x = x</span>²<span> = x * x

For the given, we want to get the square of 8. This means that we will <u>multiply 8 by itself.</u>
Therefore:
square of 8 = 8</span>²<span> = 8 * 8 = 64

Hope this helps :)</span>
8 0
3 years ago
Read 2 more answers
Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.8. (Round your ans
Alenkinab [10]

Answer:

a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 50, \sigma = 1.8

(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?

Here n = 17, s = \frac{1.8}{\sqrt{17}} = 0.4366

This probability is 1 subtracted by the pvalue of Z when X = 51. So

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

By the Central Limit Theorem

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

Z = \frac{51 - 50}{0.4366}

Z = 2.29

Z = 2.29 has a pvalue of 0.9890

1 - 0.989 = 0.011

0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?

Here n = 17, s = \frac{1.8}{\sqrt{45}} = 0.2683

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

Z = \frac{51 - 50}{0.0.2683}

Z = 3.73

Z = 3.73 has a pvalue of 0.9999

1 - 0.9999 = 0.0001

0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

8 0
3 years ago
Help and explain how you got it.<br><br> no trolling or report!
Galina-37 [17]

Answer:

I am 100% confident in my calculations. I'm not 100% sure about interpreting. Sorry I can't be of more help.

Part A:

Median: 4.5

IQR: 3.5

Mode: 6

I think the median/IQR is a better measure of center and variability because the range of data is so big (10 - 0 = 10), and the data skews to the right with the 6s.

Part B:

I would say no. It is the one that occurs most frequently, but there are many more times she rides less than 6, which is why the median is 4.5

Step-by-step explanation:

Mean: 4.714

MAD: 2.43

Median: 4.5

IQR: 3.5

Mode: 6

the number set in order:

0, 2, 2, 2.5, 2.5, 3, 4, 5, 6, 6, 6, 8, 9, 10

The median (middle number) is two numbers: 4 and 5, so the median is 4.5

The first quartile/Q1 (middle of the bottom half of data) is 2.5, the third quartile/Q3 (middle of the top half of data) is 6. So the IQR (interquartile range) is 6 - 2.5 = 3.5

The mean is the sum of the numbers divided by the quantity of the numbers, so 66/14 = 4.714

the mean absolute deviation is found by calculating the mean, and then finding the absolute value of the difference between each number of the set and the mean...then adding those together and finding the mean of that set. Labor intensive, but that's 34/14 = 2.42857143

4 0
3 years ago
What is 18.4 times 3.40 and the wprk showed
dusya [7]

Decimal division is super easy, let me show you how.

First, let's remove the decimal from both numbers. We have <em>184 </em>and <em>340</em>. Now let's multiply those two numbers.

184*340 = 62560.

Now, we need to figure out how many decimal places are in the original decimals. 18.<u>4</u> has one decimal place, and 3.<u>4</u> has one decimal place. (We don't count zeros at the end of a decimal as a decimal place.)

Now that we've figured out that there was a total of two decimal places, there can only be two decimal places in the answer. Let's take the 62560 that we got, and put a decimal in it. Remember, only two decimal places.

If there's only two decimal places in 62560, it would look like this: 62.<u>56</u> (<em>Remember, zeros at the end of a decimal does not count as a decimal place!)</em>


There's your final answer. 62.56.


Hope this explanation helped you out. :)

7 0
3 years ago
a wrestler is weighing in for their match this evening. their normal weight is 168 pounds, but that weight can vary by as much a
Natalija [7]

<em>Introduction </em>

<em>Wrestling is a wonderful activity with many advantages for the student-athlete. It is a sport that is highly </em>

<em>competitive, exciting and satisfying. It is a sport that provides for individual and team competition. It is - </em>

<em>and should be - fun. Unfortunately, the practice of losing weight by not eating, restricting fluid intake and </em>

<em>over-exercising reduces the sport's fun. This information is presented to help clear up misconceptions </em>

<em>regarding wrestling and weight loss. I also hope to give some guidance to those who desire to manage </em>

<em>their weight properly in preparation for and during the wrestling season. </em>

<em>History and Stigma </em>

<em>For too long, the wrestling community has unthinkingly accepted the myth that to be a good wrestler, you </em>

<em>must cut weight. The generally accepted thinking is something like this: if your natural weight is 135 </em>

<em>pounds, you may be a good wrestler at 135 pounds. But if you wrestle at 130 pounds, you'll be a better </em>

<em>wrestler. And if you can make it down to 125, you'll be a state champion. No facts support that widely held </em>

<em>view, yet wrestlers and parents subscribe to that faulty reasoning. Looking further back, many remember </em>

<em>the days that losing excessive weight was a specific practice and expectation among wrestlers. It was </em>

<em>supposed to teach sacrifice, commitment, and the idea of “No Pain, No Gain.” </em>

<em>Regardless of the current attitude of the majority of the wrestling community, the stigma that an unhealthy </em>

<em>loss of weight is a requirement of wrestling among outside observers sticks. The Virginia High School </em>

<em>League has followed national guidelines to establish rules that encourage healthy weight management </em>

<em>among wrestlers. </em>

<em>Current Regulations </em>

<em>Preseason weight certification is accomplished with three steps. The first is determining each wrestler’s </em>

<em>body fat percentage using skin fold calipers. Next, the wrestler completes a hydration test, to insure </em>

<em>against a dehydrated weight measurement, and the wrestler weighs in. The third step is the calculation of </em>

<em>the wrestler’s minimum wrestling weight based on 7% body fat for males and 12% for females. The </em>

<em>wrestler may not wrestle at a weight class below his minimum weight during the season. The wrestler is </em>

<em>also restricted from losing more then 1.5% body weight loss per week (official weigh-ins at certification, </em>

<em>matches, and tournaments are used for this calculation). A one-pound per month growth allowance is </em>

<em>provided to allow for the natural growth of this age group. The VHSL advises, and we have instituted, </em>

<em>daily weigh-ins before and after practice to monitor weight-loss and dehydration. </em>

<em>Additionally, the National Federation of State High School Associations has adopted several other rules to </em>

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5 0
3 years ago
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