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tino4ka555 [31]
2 years ago
10

The ages of the members of the Chess Club and the Debate Team at Jana’s school are shown in the box plots. Which statement is tr

ue regarding these data sets?
Mathematics
2 answers:
V125BC [204]2 years ago
8 0

Answer:

C lul

Step-by-step explanation:

brainly.com/question/22583943

(The majority of the members in each club are between 15 and 17 years old.)

Fynjy0 [20]2 years ago
3 0
C lol lol I need points I’m so sorry
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Find the perimeter of the following polygon. Be sure to include the correct unit in your answer.
Natalija [7]

Answer:

The answer is 48yd.

Step-by-step explanation:

All you have to do is add up the numbers.

3 0
3 years ago
Find the simple interest on a loan of 23,000 at 5.4% interest for 11 months
adell [148]

Answer:

Simple interest=1138.5

Step-by-step explanation:

Simple interest=23000×(5.4/100)×(11/12)

S.I=23000×0.054×(11/12)

S.I=1138.5

3 0
3 years ago
In a college of exactly 2660 students, exactly 60 % are male. what is the number of female students?
garik1379 [7]
If 60% out of the 2,660 students are males, then
40% out of t<span>he same 2,660 students are females

Ten the number of females is 40% x 2660 = 0.4 x 2660 = 1,064 females</span>
5 0
3 years ago
FIND g rounded to the nearest degree ??
Harrizon [31]
In this situation I would use cosine to find m∠G.

Cosine = Adjacent side ÷ Hypotenuse

RG being the adjacent side and TG being the hypotenuse, you would set up the equation like this

cos(g) = 8/17

then solve

cos(g) = 0.47058

simplify

g = cos^-1(0.47058)

and you get

m∠G = 61.9°


So your answer would be 62°
4 0
3 years ago
Read 2 more answers
A particular fruitâs weights are normally distributed, with a mean of 598 grams and a standard deviation of 6 grams. The heavies
ruslelena [56]

Answer:

The 8% of the fruit weigh more than x=606.43 \  g

Step-by-step explanation:

From the question we are told that

  The mean is  \mu = 598 \ g

    The standard deviation is \sigma = 6 \  g

Generally the 8% is mathematically represented as

    P(X >  x) = 0.08

=>   P(X >  x) = P ( \frac{X - \mu}{\sigma }>\frac{x - 598}{6}  )=0.08

\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )

       P(X >  x) = P ( Z >\frac{x - 598}{6}  )=0.08

From the normal distribution table the critical value corresponding  area representing 0.08 towards the right tail of the curve is

       z = 1.405

So

       \frac{x- 598}{6}  = 1.405

=> x=606.43 \  g

     

6 0
3 years ago
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