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3 years ago

6

On a piece of paper, graph y+ 2

le" class="latex-formula"> 1/4 x-1. Then determine which answer choice matches the graph you drew.

1 answer:

xxMikexx [17]3 years ago

7 0

Answer:

Graph A

Step-by-step explanation:

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Helppppppppppppppp lol

Sladkaya [172]

The given expression is:
(x/2) + 5

Let's look at each terms separately:
x/2 means that you you're getting half the value of a number
+5 means that you are adding 5
Combining both, you can conclude that you are adding 5 to half a certain given number.

Comparing this with the choices, the correct choice would be:
C. five more than half a number.

5 0

3 years ago

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5. Two unbiased dice are rolled. Calculate the probability that the sum of the two dice is:

IceJOKER [234]

The sample space has 36 possible pairs from 1,1 1,2 1,3 up to 6,5 and 6,6

(a). Three pairs add to 4 1,3 2,2 and 3,1 so P(4) = 3/36 = 1/12

(b). 6 pairs add to 7 so P(7) = 6/36 = 1/6

(c) 15 pairs add to less than 7 so P(<7) = 15/36 = 5/12

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2 years ago

The graph below shows the solution set of which inequality ?? HELP FASTTT

Maru [420]

<u>Answer:</u>

<h2>A.</h2>

The graph shows the solution set of the equality √x < -1

<em>(No Solutions)</em>

4 0

2 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago

If I make $7.25 a hour and I work for three days how much money is that every two weeks

trasher [3.6K]

There are a couple of things you didn't mention.
In order to come up with an answer to the question,
I have to assume reasonable numbers for them, and
then answer the question that I have invented.

Assumption #1: Each day that you work, you work for 8 hours.

Assumption #2: You work for three days each week.

If those assumptions are true, then you earn

($7.25 / hour) x (8 hours/day) x (3 days/week) x (2 weeks) =

($7.25 x 8 x 3 x 2) = $348 before taxes and other deductions.

3 0

3 years ago

Read 2 more answers