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irinina [24]
2 years ago
13

There are 8 sophomores on the academic team. At the last competition, they each took the math test. Their scores were 82%,

Mathematics
2 answers:
AysviL [449]2 years ago
7 0

Answer:

79%

Step-by-step explanation:

<em><u>m</u></em><em><u>e</u></em><em><u>d</u></em><em><u>i</u></em><em><u>a</u></em><em><u>n</u></em><em><u> </u></em>is the middle no.

arrange the scores in descending order

72,74,76,(78,80),82,92,92

78+80, divide 2= 79%

<h3 />
bearhunter [10]2 years ago
6 0

It Is A.) On Edge, Have Fun Dreamers!

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The mean number of students in a classroom at school a is 32.5 and there are 25 classrooms. The mean number of students at schoo
Airida [17]

We have given three school there, School A, School B and School C.

We are going to use following formula.

Total number of student in a school = Mean of number of student × Number of classrooms.

School A

Mean of students in School A = 32.5.

Number of classroom in School A = 25.

Number of students in School A = Mean × number of classrooms = 32.5×25 = 812.5


School B

Mean of students in School B = 29.6.

Number of classroom in School B = 12.

Number of students in School B = Mean × number of classrooms = 29.6×12 = 355.2


School C

Mean of students in School C = 15.3.

Number of classroom in School C = 10.

Number of students in School C = Mean × number of classrooms = 15.3×10 = 153.


Total sum of number of classes in all schools = 25+12+10 = 47.

Total sum of all students in all schools = 812.5 +355.2 +153 = 1320.7 .

The mean number of students per classroom for all the schools combined =

\frac{Total \ sum \ of \ all \ students \ in \ all \ schools}{Total \ sum \ of \ number \ of \ classes \ in \ all \ schools}

= \frac{1320.7}{47} = 28.1

The mean number of students per classroom for all the schools combined= 28.1

4 0
2 years ago
If lmn xyz which congruences are true cpctc check all that apply i.will give you brainiest
bixtya [17]
A
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Those are your answers.
5 0
2 years ago
Read 2 more answers
Lisa has $150 at most to spend on clothes. She wants to buy a pair of jeans for $58 and will spend the rest on t-shirts that cos
anzhelika [568]

Answer:

14t + 58 ≤ 150

Step-by-step explanation:

If she cannot spend more than what she has, which is 150, the inequality sign has to be "less than or equal to".  It's ok if she spends less than 150, but not ok if she spends more, because she doesn't have it to spend.

We know the cost of 1 pair of jeans is 58.  Now she wants to make up the difference by getting as many $14 shirts as possible (the number of shirts being our unknown).  

That means that the cost of the jeans PLUS the unknown number of shirts cannot exceed 150.

Therefore, the inequality is:

14t + 58 ≤ 150

4 0
2 years ago
In a network of 40 computers, 5 hold a copy of a particular file. Suppose that 7 computers at random fail. Let F denote the numb
anzhelika [568]

Answer:

a. E(F)=0.875

b. 99.9976%

c. P(X=2)=0.1683

Step-by-step explanation:

a. We notice that this is a binomial distribution with the probability of success;

p=\frac{5}{40}=0.125

#We are given the sample size, n=7. The Expected value is calculated as:

E(X)=np\\\\E(F)=np , n=7, p=0.125\\\\E(F)=7\times 0.125\\\\=0.875

Hence the expectation, E(F)=0.875

b. To calculate the probability of the range of F, we need to calculate all possible outcomes of F in the given sample;

P(X\leq 5)=1-P(X=6)-P(X=7)\\\\=1-{7\choose 6}(0.125)^6(1-0.125)^1-{7\choose 7}(0.125)^7(1-0.125)^0\\\\=1-0.000023365-0.000000476\\\\=0.999976158\\\\=99.9976\%

Hence, the range of F is 99.9976%

c. The probability that F=2 is calculated  using the binomial distribution function as:

P(X=2)={7\choose 2}(0.125)^2(1-0.125)^5\\\\=0.1683

Hence, the probability of F=2 is 0.1683

3 0
2 years ago
A 12-foot high ladder reaches a tree limb when it leans at a 36 degree angle with the ground. Approximately how far away is the
Allisa [31]
Look at the picture.

\cos 36^\circ = \frac{x}{12} \\&#10;0.809 \approx \frac{x}{12} \\&#10;0.809 \times 12 \approx x \\&#10;x \approx 9.708

The base of the ladder is approximately 9.708 feet from the tree trunk base.

4 0
2 years ago
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