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3 years ago

How many times does 13 go into 80

Mathematics

1 answer:

Tcecarenko [31]3 years ago

6 0

13 x 6 = 78 you cant multiply anymore so that's how many times it can go in 80.

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The distance between the points (3,y1) and (11,11) is 17. What is a possible value of y1?

34kurt

C I’m pretty sure hope this helps

4 0

2 years ago

Corinne has a job selling magazines. She earns $7.50 per hour plus 20% of the total amount of her sales. She also gets an allowa

ra1l [238]

Earnings=20% of sales+7.5h+40

coeficinets are 7.5 and 0.2
vraiables are h and s
constant is 40

B. if h=25 and s=30 input,
7.5*25+0.2*300+40=$287.50

C
new equation is 9h+0.2s+40
coefienct woud be 9 and 0.2
variable is h and s
constant is 4

7 0

3 years ago

What is the value of the correlation coefficient r of the data set?

enyata [817]

The correct option will be: C) 0.84

Explanation

Formula for Correlation coefficient :

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

First, for each point(x, y), we need to calculate x², y² and xy .

Then, we will find sum all x, y, x², y² and xy, which gives us Σx, Σy, Σx², ∑y² and Σxy

(Please refer to the attached image for the table )

Here we got, ∑x = 44 , ∑y = 183 , ∑x² = 362 , ∑y² = 6575 and ∑xy = 1480

'n' is the total number of data set, which is 7 here.

So, plugging those values into the above formula..........

$r= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}}\\ \\ r=\frac{7(1480)-(44)(183)}{\sqrt{[7(362)-(44)^2][7(6575)-(183)^2]}}\\ \\ r=\frac{10360-8052}{\sqrt{(598)(12536)}}\\ \\ r=\frac{2308}{\sqrt{7496528}}\\ \\ r=0.84$

So, the value of the correlation coefficient is 0.84

4 0

3 years ago

Read 2 more answers

The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.

5 0

3 years ago

What is the probability that a family of 4 children will have all girls?

Alex73 [517]

Since as of now we're saying that the probability of having a girl is 1/2, we can say:
1/2 * 1/2 * 1/2 * 1/2, because then you're saying a one half chance of having a girl times another one half chance of having a girl, etc.
That ends up to be 1/16.

3 0

3 years ago