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dolphi86 [110]
3 years ago
7

on the last 5 quizzes, marco has earned a total of 21 out of 25 possible points. if he has a perfect score on the next several q

uizzes, exactly how many points will he need to bring up his average up to 90%
Mathematics
1 answer:
nevsk [136]3 years ago
6 0

Using the mean concept, it is found that we will need 15 more points to bring up his average up to 90%.

The mean of a data-set is the <u>sum of all observations divided by the number of observations</u>.

In this problem:

  • In the first 5 observations, total of 21 out of 25 points, hence the mean for these observations is \frac{21}{25} = 0.84.
  • In the next n observations, mean of 1.

Hence, the mean is:

M = \frac{0.84(5) + n}{5 + n} = \frac{4.2 + n}{5 + n}

We want the mean to be of 0.9, thus:

M = 0.9

\frac{4.2 + n}{5 + n} = 0.9

4.2 + n = 0.9n + 4.5

0.1n = 0.3

n = \frac{0.3}{0.1}

n = 3

3 more testes are need, each worth 5 points, hence, 15 more points are needed to bring up his average up to 90%.

A similar problem is given at brainly.com/question/25323941

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Whoops i forgot to read the lesson :P 30 points
andriy [413]

Step-by-step explanation:

I'll do 2.

Alright,Alex let say we have factored a quadratic into two binomial, for example

(5x + 3)(6x - 4)

If we set both of those equal to zero

(5x + 3)(6x + 4) = 0

We can used the zero product property in this case to find the roots of the quadratic equation.

This means that

if  \: ab = 0 \: then \: a = 0 \: and \: b = 0

This means we set each binomal equal to zero to find it root.

5x + 3 = 0

5x =  - 3

x =  -  \frac{3}{5}

6x + 4 = 0

6x =  - 4

x =  -  \frac{2}{3}

So our roots are negative 3/5 and negative 2/3 using zero product property

6 0
3 years ago
A beacon is flashing on top of a 50 foot tower. A 6 foot tall man walks constantly away from the tower at 5 feet/sec. At the ins
Viefleur [7K]

Answer:\frac{253}{44}

Step-by-step explanation:

ignore the "at the instant the man is 30 feet away" part, set it as X and the man's shadow as Y.

Similar triangles so we can do \frac{50}{x+y}  = \frac{6}{y}.

Solve for it we get 44y = 6x

Differentiate relative to time t, we get 44y' = 6x'.

change in x (x') is equal to 5. And we get the answer y' = \frac{33}{44}.

the \frac{33}{44} ft/sec is the rate of which the length of the shadow is changing. add 5 to it for the rate of the tip of his shadow moving away from the tower.

7 0
2 years ago
Need help<br>here is the question and the answers​
scoundrel [369]

take a picture of the display counter too

3 0
3 years ago
Help please asap algebra 2
timurjin [86]
X = approximately 633

Steps:
lnx + ln3x = 14

ln3x^2 = 14 : Use the log property of addition which is to multiply same log                           together so you multiply x and 3x because they have log in                                common

  (ln3x^2) =  (14)  : take base of e on both sides to get rid of the log
e                e

3x^2 = e^14  : e cancels out log on the left side and the right side is e^14 

x^2 = e^14 / 3 : divide both sides by 3

√x^2 = <span>√(e^14 / 3) : take square root on both sides to get rid of the square 2                                  on x
</span>
x = √(e^14) / <span>√3 : square root cancels out square 2 leaving x by itself

x = e^7 / </span>√3 : simplify the √(e^14) so 14 (e^14) divide by 2 (square root) = 7<span>

x = </span>633.141449221 : solve 
4 0
3 years ago
21) Bob ran the first part of a 12 km race at a speed of 8 kmph. He ran the second part of the
diamong [38]

Answer:

Step-by-step explanation:

Bob ran the first part of a 12 km race at a speed of 8 kmph. He ran the second part of the

race at 10 kmph. If his total time for the entire race was 1.74 hours, how far did he run in

the first part of the race?

First part :

Total Distance = 12 km

First part Speed = 8kmph

Let distance covered on first part = x

Second part speed = 10kmhr

(8*x)

6 0
3 years ago
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