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german
3 years ago
10

Harold wants to get at least a grade of 84 in math class. His grade will be the average of two test scores. He got a 77 on the f

irst test.
A. Write an inequality to represent the situation.


B. What is the least grade Harold needs to get on his second test to mee his goal?

This is my last question for my homework! Please help me out :)
Mathematics
2 answers:
Andrew [12]3 years ago
4 0
1. (77+x)/2=84

2. Harold needs at least a 91

Work:

first, you multiply both sides by 2; 77+x=168
then you subtract 77 from both sides; x=91
then you plug it back in to check; (77+91)/2=84 ; 168/2=84 ; 84=84
balu736 [363]3 years ago
4 0

Answer:

\frac{77+x}{2} \geq 84

Harold needs to get more than 91  on his second test

Step-by-step explanation:

Harold wants to get average of Two test grades in math class is atleast 84.

He got 77 on the first test. and Let x be the score of second test.

Average of two test is \frac{77+x}{2}

Average is atleast 84. The inequality becomes

\frac{77+x}{2} \geq 84

Now solve the inequality for x

\frac{77+x}{2} \geq 84

Multiply both sides by 2

77+x\geq 168

Subtract 77 from both sides

x\geq 91

Harold needs to get more than 91  on his second test

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If for a particular one-to-one function f(2) = 13, what are the corresponding input and output
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3 years ago
Nina and Jo both ran an 8 km race. Nina took 55 minutes to run the whole race. Jo started the race 3 minutes later than Nina but
grin007 [14]

Answer:

Step-by-step explanation:

First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is

\frac{8km}{55min}=.145\frac{km}{min} and we can use that in a d = rt table:

                 d        =        r        *        t

Nina                            .145

Jo

Now we can fill in the distance which is 6 for both, since that is the distance where they met:

                d        =        r       *        t

Nina         6        =     .145

Jo             6        =

Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:

                d        =        r        *        t

Nina          6       =       .145    *      t + 3

Jo              6       =         r       *         t

As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then

6 = .145(t + 3) and

6 = .145t + .435 and

5.55 = .145t so

t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then

6 = r(38.379) and

r = .16 km/min

Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:

6 should equal .145454545( 38.37931034 + 3)

6 ?=? .145454545(41.37931034)

6 ?=? 6.0 so

Jo's rate is .16 km/min rounded

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