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Tems11 [23]
2 years ago
5

The scatterplot shows the ages and finishing times of seven men who ran a charity 5K run. Use the labeled points to create a lin

ear model that predicts the race time (y) of a man of any age (x). Which equation represents this linear model?
a.y=0.47+ 7.72
b.y=0.47+29.15
c.2.146X-27.57
d.2.14X-2.07

Mathematics
2 answers:
kondor19780726 [428]2 years ago
8 0
The correct option is (a) <span>y=0.47x+ 7.72
</span>
Explanation:
The two points given are:
(21, 17.5) and (43, 27.75)

First you need to find the slope:
Slope = (27.75-17.5)/(43-21) = 0.466

Now equation of line passing through the point (21, 17.5) and having slope 0.466 is:

y - 17.5 = 0.466(x - 21)
y - 17.5 = 0.466x - 9.78
y = 0.47x + 7.72
Savatey [412]2 years ago
8 0

Answer:

The right answer is A

Step-by-step explanation:

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Find the length of the leg of a right triangle with leg length b= 21.5 inches and the hypotenuse c= 31.9 inches. Use a calculato
sweet [91]

Answer:

23.6 (approximate value)

Step-by-step explanation:

  • a*a+b*b=c*c Pythagorean thereom
  • (21.5)^2 + b*b = (31.9)^2
  • b*b= 1017.61 - 462.25
  • √b*b= √555.36
  • b=23.6(approximate value)
8 0
2 years ago
Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
2 years ago
If 2x2 + y2 = 17 then evaluate the second derivative of y with respect to x when x = 2 and y = 3. Round your answer to 2 decimal
Vera_Pavlovna [14]

Answer:

y''=-1.26

Step-by-step explanation:

We are given that 2x^2+y^2=17

We have evaluate the second order derivative of y w.r.t. x when x=2 and y=3.

Differentiate w.r.t x

Then , we get

4x+2yy'=0

2x+yy'=0

yy'=-2x

y'=-\frac{2x}{y}

Again differentiate w.r.t.x

Then , we get

2+(y')^2+yy''=0 (u\cdot v)'=u'v+v'u)

2+(y')^2+yy''=0

Using value of y'

yy''=-2-(-\frac{2x}{y})^2

y''=-\frac{2+(-\frac{2x}{y})^2}{y}

Substitute x=2 and y=3

Then, we get y''=-\frac{2+(\frac{4}{3})^2}{3}

y''=-\frac{18+16}{9\times 3}=-\frac{34}{27}

Hence,y''=-1.26

6 0
2 years ago
Why did the queen have the king measure the rug
Alex73 [517]
Because he was such a good ruler
5 0
2 years ago
Max travels to see his brother's family by car. He drives 216 miles in 4 hours.
LUCKY_DIMON [66]

Answer:

54 mph

Step-by-step explanation:

3 0
2 years ago
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