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sergeinik [125]

3 years ago

10

On Monday, it snowed a total of 15 inches. On Tuesday and Wednesday, it snowed an additional 4 1/2 inches and 6 3/4, respectivel

y. A weather forecaster says that over the last three days, it snowed over 2 1/2 feet . Is this a valid claim? Justify your answer

1 answer:

Serggg [28]3 years ago

5 0

How were u taught to do it

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25 is how much more than 20? what percent more?

Firdavs [7]

25 is 5 more than 20 which is 25% more since 5/20 = 1/4 = 25%
To check (1.25 * 20 = 25)

7 0

3 years ago

Read 2 more answers

The correlation between the height and weight of children aged 6 to 9 is found to be about r = 0.8. Suppose we use the height x

Andrew [12]

Answer:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The value of r is always between $-1 \leq r \leq 1$

And we have another measure related to the correlation coefficient called the R square and this value measures the % of variance explained between the two variables of interest, and for this case we have:

$r^2 = 0.8^2 = 0.64$

So then the best conclusion for this case would be:

c. the fraction of variation in weights explained by the least-squares regression line of weight on height is 0.64.

Step-by-step explanation:

For this case we know that the correlation between the height and weight of children aged 6 to 9 is found to be about r = 0.8. And we know that we use the height x of a child to predict the weight y of the child

We need to rememeber that the correlation is a measure of dispersion of the data and is given by this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The value of r is always between $-1 \leq r \leq 1$

And we have another measure related to the correlation coefficient called the R square and this value measures the % of variance explained between the two variables of interest, and for this case we have:

$r^2 = 0.8^2 = 0.64$

So then the best conclusion for this case would be:

c. the fraction of variation in weights explained by the least-squares regression line of weight on height is 0.64.

3 0

3 years ago

Tony is building a doghouse and the front

Viktor [21]

Answer:

and the front what can you post more or is that the question if so good for him to build his dog a house

Step-by-step explanation:

6 0

3 years ago

What is the equation of this line in slope-intercept form?

Elis [28]

Answer:

y=-1/2x+2

Step-by-step explanation:

for x-intercept

put y=0

0= -1/2x+2

1/2x=2

x=4

(4,0)

for y-intercept

put x=0

y= 2

(0,2)

7 0

3 years ago

Read 2 more answers

You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

$f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.$

a) We calculate the probability that you need less than 5 minutes to get up:

$P(x$

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate the probability that you will need between 8 and 13 minutes:

$P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3$

Therefore, the probability is P(8<x<13)=0.3.

d) We calculate the probability that you will be late to each of the 9:30am classes next week:

$P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6$

You have 9:30am classes three times a week. So, we get:

$P=0.6^3=0.216$

Therefore, the probability is P=0.216.

e) We calculate the probability that you are late to at least one 9am class next week:

$P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1$

Therefore, the probability is P=1.

3 0

3 years ago