Ask question

Ask question

All categories

2 years ago

13

Fred is playing tennis. For every 2 of his serves that land in, 4 serves land out. If he hit 28 serves in, how many serves lande

d out?

2 answers:

bija089 [108]2 years ago

5 0

2 for every 4 = 1 for every two, so 28 multiplied by two. Your answer is 56.

ZanzabumX [31]2 years ago

4 0

56 serves were landed out

You might be interested in

A triangle has side lengths of 7, 24, and 25. Is the triangle right, acute, or obtuse?

SOVA2 [1]

I think that it's an obtuse triangle.

6 0

2 years ago

Read 2 more answers

PLEASE HELP WITH THIS QUESTION

Radda [10]

Answer:

The correct answer is Redbirds

P = 4/5 = 0.8

4 0

2 years ago

Evaluate the expression 3^3+6(2+3/6)

fiasKO [112]

42

because 27+6(2.5)= 27 +15 = 42

8 0

3 years ago

Read 2 more answers

25% of American households have only dogs (one or more dogs) 15% of American households have only cats (one or more cats) 10% of

sergeinik [125]

Answer:

a) P=0.2503

b) P=0.2759

c) P=0.3874

d) P=0.2051

Step-by-step explanation:

We have this information:

25% of American households have only dogs (one or more dogs)

15% of American households have only cats (one or more cats)

10% of American households have dogs and cats (one or more of each)

50% of American households do not have any dogs or cats.

The sample is n=10

a) Probability that exactly 3 have only dogs (p=0.25)

$P(x=3)=\binom{10}{3}0.25^30.75^7=120*0.01563*0.13348=0.25028$

b) Probability that exactly 2 has only cats (p=0.15)

$P(x=2)=\binom{10}{2}0.15^20.85^8=45*0.0225*0.27249=0.2759$

c) Probability that exactly 1 has cats and dogs (p=0.1)

$P(x=1)=\binom{10}{1}0.10^10.90^0=10*0.1*0.38742=0.38742$

d) Probability that exactly 4 has neither cats or dogs (p=0.5)

$P(x=4)=\binom{10}{4}0.50^40.50^6=210*0.0625*0.01563=0.20508$

8 0

3 years ago

40% of what number is 80

Arte-miy333 [17]

40%*2=80 I hope I helped!

8 0

2 years ago