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klasskru [66]
2 years ago
13

Player a throws the ball 25 yards to player b. After catching the ball, player be immediately turns 90° to his left and runs wit

h the ball for 30 yards. At the end of the play, how far away is the ball from where it started? Round your answer to the nearest whole number
Mathematics
1 answer:
melamori03 [73]2 years ago
3 0

Answer: The ball is 39 yards away from where it started

Step-by-step explanation:

You might be interested in
C=.41+.26(z-1) solve for z
Ksivusya [100]

c=0.41+0.26(z-1) \\0.26(z-1)=c-0.41 \\z-1=(c-0.41)/0.26 \\\boxed{z=(c-0.41)/0.26+1}

Hope this helps.

6 0
3 years ago
64. Denise wants to burn at least 5000 calories a week through running. Based on her running speed, she estimates that she can b
Rudiy27

An inequality that represents Denise's goal in terms of the number of hours spent running 'h'

550h\leq 5000

Given :

Denise wants to burn at least 5000 calories a week through running.

she can burn 550 calories per hour

Let 'h' be the number of hours spent

In 1 hour  she can burn  550 calories

In 'h' hours she can burn 550h calories

Given that she want to burn atleast 5000 calories in a week

Burn atleast 5000 calories means <=5000 calories

So , the inequality that represents Denise's goal is

calories burn in h hours  <= 5000 calories

550h\leq 5000

Learn more :  brainly.com/question/381815

5 0
2 years ago
Plzzzzzzzzzzzzzzzzzz
sveta [45]

The first column. The rule for that column is y = x * -4.

8 0
3 years ago
What is the slope of the equation y = 3/4x + 8?<br>A. 3/4<br>B. 8<br>C. 3<br>D. 4<br>​
steposvetlana [31]

Answer:

Step-by-step explanation:    

y = m x + b

y = 3 /4 x − 8

Rewrite in slope-intercept form.

y = 3 /4 x − 8

Using the slope-intercept form, the slope is  3 /4 .

m =3 /4

6 0
3 years ago
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

8 0
3 years ago
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