Ask question

Ask question

All categories

3 years ago

13

I am confused by this can someone help me, please! Everything is on the photo.

2 answers:

Marina CMI [18]3 years ago

6 0

Thank you for posting your question on Brainly.

E) a^4 = Some number to the power of 4
D) 4x = Four times an unknown value
C) z ÷ 4 = Quotient of a variable and four
B) y - 4 = Four less than an unknown value
A) b + 4 = Four more than some number

I hope my answer is satisfactory to you. Thank you.

Free_Kalibri [48]3 years ago

5 0

It's trivial..

1. B
2. C
3. E
4. D
5. A

You might be interested in

Use the table to determine the average rate of change from 3 to 6 storms

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Good Morning If you can be more specific i can help you with this question but there is no table text me back if oyu still need help! ;)

5 0

3 years ago

Tristan and two of his friends purchased tickets to a school play. Tristan paid for all the tickets with a $50.00 bill and recei

Elza [17]

Answer: 27.00

Step-by-step explanation:

$50 - $23 = $27

3 0

3 years ago

Read 2 more answers

(44+6x) / (16+x) ………

Anastasy [175]

The answer would be 6- 52/(x+16)

3 0

2 years ago

Read 2 more answers

Can someone tell me if I'm correct or not?

Gnoma [55]

No the vertex should be at (-3,-2) (I checked on my calculator)

5 0

3 years ago

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