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TiliK225 [7]
3 years ago
6

Hours Rented (h) Cost ($) 1. Determine whether the relationship between the two quantities described in the table is linear. If

so, find the constant rate of change. If not, explain your reasoning. 2 50 1. 4 100 6 150 8 200 Is the relationship linear? (type (type YES or NO) What is the constant rate of change (slope)?​
Mathematics
1 answer:
Pavel [41]3 years ago
6 0

Answer: I think it is either 25$ an hour or no the  change from cost to hours is not constant

Step-by-step explanation:

Hope that helpssss !!!!!!!!!!! *-*

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4 (x+2) = 2x+10<br> help
xenn [34]

Answer:

x=1

Step-by-step explanation:

First you would do the parentheses.

So, 4 x X= 4x

4x2=8

4x+8=2x+10

subtract 2x on both sides

2x+8=10

subtract 8 from both sides

2x=2

x=1

6 0
3 years ago
If you add Natalie's age and Fred's age, the result is 42. If you add Fred's age to 3 times
Alla [95]

Answer:

N+F=42 & F+3N=70 System Solved: Natalie is 14 and Fred is 28

Step-by-step explanation:

Using variables N=natalies age and F=Freds age

N+F=42 ----> N=42-F (take this and plug it into the other equation)

F+3N=70 so F+3(42-F)=70 [simplify] F+126-3F=70 [further simplify]

-2F=-56 [divide -56 by -2] F=28 [then plug this number into the orig. equation]

So N+28=42 or N=14.

5 0
3 years ago
Suppose that 10% of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeown
guapka [62]

Remainder of question:

Find the probability distribution of x

Answer:

The random variable x is defined as: X = {0, 1, 2, 3, 4}

The probability distribution of X:

P(X = 0) = 0.656

P(X = 1) = 0.2916

P(X= 2) = 0.0486

P(X=3) = 0.0036

P(X = 4) = 0.0001

Step-by-step explanation:

Sample size, n = 4

Random variable, X = {0, 1, 2, 3, 4}

10% (0.1) of the homeowners are insured against earthquake, p = 0.1

Proportion of homeowners who are not insured against earthquake, q = 1 - 0.1

q = 0.9

Probability distribution of x,

P(X = r) = ^nC_r *p^r q^{n-r} \\\\P(X= 0) =(^4C_0 *p^1 q^4 )\\P(X=0) = (^4C_0 *0.1^0 0.9^4 ) = 0.656\\P(X= 1)= (^4C_1 *p^1 q^3 )\\P(X=1) = (^4C_1 *0.1^1 0.9^3 ) = 0.2916\\P(X= 2)=( ^4C_2 *p^2 q^2) \\P(X=2) = (^4C_2 *0.1^2 0.9^2 ) = 0.0486\\P(X= 3) = (^4C_3 *p^3 q^3) \\ P(X=3) = (^4C_3 *0.1^3 0.9^1 ) = 0.0036\\P(X= 4) =  (^4C_4 *p^4 q^0 )\\ P(X=4) =(^4C_4 *0.1^4 0.9^0 ) = 0.0001

5 0
3 years ago
What is the answer to all of them?
melisa1 [442]
The answers are
x=5
n=3
y=4
k=-80
3 0
3 years ago
I don’t know how to do this. Ignore my work it’s probably wrong. Please help.
wariber [46]

Check the picture below.

A)

well, we start off by making a table of values for the function, with a few "t" values, as you see in the picture, then graph those points to get the graph.

B)

at t = 0, h(0) = 98, and at t = 1, h(1) = 82, so it went from 98 to 82, for a difference of 16.

C)

at t = 1, h(1) = 82, and at t = 2 h(2) = 34, so it went from 82 to 34, for a difference of 48.

does it fall the same distance on both intervals? well, they're different, so nope.

6 0
3 years ago
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