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erastova [34]
3 years ago
14

On a map where each unit represents one kilometer, two marinas are located at P(4,2) and Q(8,12). If a boat travels in a straigh

t line from one marina to the other, how far does the boat travel?
A.

B.

C.

D.



Mathematics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

B

Step-by-step explanation:

draw a right angled triangle with the right angle in the bottom left

label P as the bottome left corner and write its coordinates next to it

do same with Q

the length of the triangle is 4 (8-4 = 4)

the height is 10 (12-2 = 10)

use pythagoras to work out the hypotenuse:

\sqrt{4^{2}+10^{2}} =2\sqrt{29}

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Algebra 2: answer these three questions. all of the questions are in the picture below
Kisachek [45]

1)

i) Reflection about y axis i.e. x=0

ii) Horizontal shrink by scale factor 4

iii) vertical shift by 4 units up.

-----------------------------------

2) f(x) passes through (1,-1) and (0,1)

So slope = (1+1)/(0-1) = -2

Using point slope form equation of f(x) is

y-1 = -2 (x-0) or y = -2x+1

g(x) passes through (0,-1) and (1,1)

Slope = (2/1) = 2

So using point slope form g(x) is

y-1=2x  

Hence we find that y-1 = -2x is transformed into y-1 = 2x

i.e. there is a reflection of f(x) on the line y =1

-------------------------------


8 0
3 years ago
2.00x+2.40y≤48.00 <br> what equation would you get from this?
Olin [163]
Bdbsbdbdue the correct
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Witch of the following is a true statement about a circle inscribed in a regular polygon
ss7ja [257]
The area of the circle is equal to the area of the polygon
4 0
4 years ago
Which figures can be precisely defined by using only undefined terms? Select three options.
love history [14]

This is a pretty bad question but I think the answer they're looking for is

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8 0
3 years ago
Read 2 more answers
18. A normal population has a mean of 80.0 and a standard deviation of 14.0. a. Compute the probability of a value between 75.0
mixer [17]

Answer:

a) 40.17% probability of a value between 75.0 and 90.0.

b) 35.94% probability of a value 75.0 or less.

c) 20.22% probability of a value between 55.0 and 70.0.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 80, \sigma = 14

a. Compute the probability of a value between 75.0 and 90.0.

This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 75.

X = 90

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

Z = \frac{90 - 80}{14}

Z = 0.71

Z = 0.71 has a pvalue of 0.7611

X = 75

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

Z = \frac{75 - 80}{14}

Z = -0.36

Z = -0.36 has a pvalue of 0.3594

0.7611 - 0.3594 = 0.4017

40.17% probability of a value between 75.0 and 90.0.

b. Compute the probability of a value 75.0 or less.

This is the pvalue of Z when X = 75. So

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

Z = \frac{75 - 80}{14}

Z = -0.36

Z = -0.36 has a pvalue of 0.3594

35.94% probability of a value 75.0 or less.

c. Compute the probability of a value between 55.0 and 70.0.

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 55.

X = 70

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

Z = \frac{70 - 80}{14}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

X = 55

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

Z = \frac{55 - 80}{14}

Z = -1.79

Z = -1.791 has a pvalue of 0.0367

0.2389 - 0.0367 = 0.2022

20.22% probability of a value between 55.0 and 70.0.

6 0
4 years ago
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