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3 years ago

The coordinates for the vertices of a right triangle are (1, 4), (6, 4), and (6,1). What is the length of the longer leg?

Mathematics

1 answer:

Nataly_w [17]3 years ago

6 0

Check the picture below.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 1 &,& 4~) 
% (c,d)
&&(~ 6 &,& 1~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(6-1)^2+(1-4)^2}\implies d=\sqrt{5^2+(-3)^2}
\\\\\\
d=\sqrt{25+9}\implies d=\sqrt{34}$

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The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c

Anestetic [448]

Answer:

a) The mean is $\mu = 60$

b) The standard deviation is $\sigma = 9$

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when $X = 55.5, Z = -0.5$

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

$-0.5 = \frac{55.5 - \mu}{\sigma}$

$-0.5\sigma = 55.5 - \mu$

$\mu = 55.5 + 0.5\sigma$

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when $X = 71.52, Z = 1.28$

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

$1.28 = \frac{71.52 - \mu}{\sigma}$

$1.28\sigma = 71.52 - \mu$

$\mu = 71.52 - 1.28\sigma$

Since we also have that $\mu = 55.5 + 0.5\sigma$

$55.5 + 0.5\sigma = 71.52 - 1.28\sigma$

$1.78\sigma = 71.52 - 55.5$

$\sigma = \frac{(71.52 - 55.5)}{1.78}$

$\sigma = 9$

$\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60$

Question

The mean is $\mu = 60$

The standard deviation is $\sigma = 9$

6 0

2 years ago

A cube. The top face has points G, B, C, F and the bottom face has points H, A, D, E.

Triss [41]

Answer:

its side CE

Step-by-step explanation:

I got it wrong and it showed me the answer. Hope this helps!

4 0

2 years ago

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Natalie uses 68 photos to make an online photo album. She can fit 5 photos on each page. She fills all of the pages in her album

Sati [7]

68-2(\5)=#of pages in album

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3 years ago

Can someone help if you can’t tell I suck at math?

Ad libitum [116K]

Answer:

Step-by-step explanation:

1. There will be 9 roses in total.

Probability Orange: 2/9

Probability Yellow: 3/9 = ⅓

Probability Pink: 4/9

2. 4/7 as there are 7 bowls in total and 4 are chocolate.

3. 4/10 = ⅖ (simplified) There are 10 fruits in total and 4 are apples.

4. 5/15 = ⅓ (simplified) There are 15 cars in total and 5 or them are cooper minis.

5. 2/6 = ⅓ (simplified) There are 6 textbooks and 2 of them are math.

6. 13/20 There are 20 drinks and 13 of them are lemonade.

7. There are 10 toys in total.

Probability toy cars: 3/10

Probability dolls: 2/10 = ⅕ (simplified)

Probability balls: 5/10 = ½ (simplified)

8. In total the number of red roses and lilies are 7. There are 10 flowers in total so the number of jasmines are 3. Therefore the probability of getting a jasmine is 3/10

9. 20/50 = ⅖ (simplified) There are 50 papers in total and 20 of them are yellow.

10. 8/18 = 4/9 (simplified) There are 18 buses and 8 of them are air conditioned.

8 0

2 years ago

(04.05 MC)

nika2105 [10]

Answer:

Step-by-step explanation:

Plug in and see.

Check A: Lets plug in (0,3) into y<2/3 x+2.

3<2/3 (0)+2

3<2 is not true so not A

Check C: Lets plug in (3,5) into y<2/3 x+2.

5<2/3 (3)+2

5<2+2

5<4 is not true so not C

Check B: Lets plug in (-3,1) into y<2/3 x+2.

1<2/3 (-3)+2

1<0 is not true so not B

Check D: Lets plug in (1,2) into y<2/3 x+2.

2<2/3 (1)+2

2<2/3+2 is true so D

6 0

2 years ago