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pantera1 [17]
3 years ago
11

a student scores 71 out of 100 on a test, if the maximum score on the tests is also 100, what score does the student need to mai

ntain at least an average of 78?
Mathematics
1 answer:
Nonamiya [84]3 years ago
8 0
(71+x)/2=78
71+x=156
x=85

Th student needs to score a min of 85
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Anyone know how to do this??
Ray Of Light [21]

Answer:

Step-by-step explanation:

Call high street is a, using  Pythagorean Theorem we have:

x^2 = 4 + a^2

b^2 = a^2 + 36

64 = x^2 + b^2

so we have x^2 = 16, a^2 = 12, b^2 = 48

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4 years ago
Which ordered pair comes with the table?
Tasya [4]
You might have the correct answer !
4 0
3 years ago
If a 12-student class averaged 90 on a test, and a 20-student class averaged 80 on the test, then all 32 students averaged
Alexxx [7]
Let the marks of the students in class 1 be

m_1, m_2, ....., m_1_2

the average of these 12 marks is 90, so 

\frac{m_1+ m_2+ .....+ m_1_2}{12}=90

which means

m_1+ m_2+ .....+ m_1_2=90*12=1080


similarly, let n_1, n_2, ....., n_2_0 be the marks of the students in the second class, 

so \frac{n_1+ n_2+ ..... n_2_0}{20}=80

which means n_1+ n_2+ ..... n_2_0=80*20=1600


The 32 students averaged:

\frac{(m_1+ m_2+ .....+ m_1_2)+(n_1+ n_2+ ..... n_2_0)}{12+20}= \frac{1080+1600}{32}= \frac{2680}{32}=   83.75
5 0
4 years ago
Read 2 more answers
Aldos coffee shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Aldo $4.15 per pound, and type of
bogdanovich [222]

Let A represent amount of Type A coffee pounds used.  

Let B represent amount of Type B coffee pounds used

A + B = 156  

B = 156 - A  

A = 156 - B

5.80A + 4.65B = 826.60  

580 (156 - B) + 4.65B - 826.60 = 0  

904.8 - 5.80B + 4.65B - 826.60 = 0

904.8 - 1.15B - 826.60 = 0  

78.2 - 1.15B = 0  

78.2/1.15 = 1.15B/1.15

68 = B

B = 68 pounds of Type B coffee

There's many more steps you can take to check and etc but am too lazy to put down sorry.

3 0
3 years ago
Katrina wants to estimate the proportion of adults who read at least 10 books last year. to do​ so, she obtains a simple random
Vaselesa [24]
E=Z*sqrt (p(1-p)/N), where E= error margin, p=proportion, N=sample size

Katrina's margin error at 85% confidence interval: E=1.96*sqrt (p(1-p)/100) = 0.196 sqrt (1(1-p))

Mathew's margin error at 99% confidence interval: E= 2.58*sqrt (p(1-p)/400) = 0.129 sqrt (p(1-p))

Since both obtained same estimate of proportion (that is, value of p), it can be seen that Mathew's estimate will have a small error (That is, 0.129 is smaller than 0.196). This can be attributed to larger sample size although a wider confidence (99%) interval was considered.
3 0
3 years ago
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