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Nookie1986 [14]
3 years ago
11

What is the value represented by the letter A on the box plot of data?

Mathematics
2 answers:
Triss [41]3 years ago
8 0

Answer:

a=5

Step-by-step explanation:

it is the lower extream

Lady_Fox [76]3 years ago
7 0
Answer: A = 5

The Wise Orange found an image that may be helpful:

You might be interested in
Twice a year, for all four years of college, you used your credit card to pay for books and supplies, which amounts to $516 a se
mixer [17]

Answer:

$11,728

Step-by-step explanation:

Twice a year, for 4 years is 8 times

516×8 = 4128

Yearly for 4 years is 4 times

700×4 = 2800

Every month for 4 years is 48 times

100×48 = 4800

Minimum expenditure:

4128 + 2800 + 4800 = $11,728

4 0
3 years ago
Wendy can wash a car twice as fast as Jason. When they work together, Wendy and Jason can wash a large van in 2 hours. How many
Ugo [173]
X= number of hours would it take Jason to wash the van by himself.
2x=number of hours would it take Wendy to wash the van by herserlf.

<span>we calcaulate the fraction of  work by Jason during one hour </span>

x hours-----------------------------1 work
1 hour----------------------   fraction of work during one hour.

fraction of work during one hour=(1 hour * 1work) / x hours=1/x


we calculate the fraction of work by Wendy during one hour.
2x hours-----------------------------1 work
1 hour----------------------   fraction of work during one hour.

fraction of work during one hour=(1 hour * 1work) / 2x hours=1/2x

We can suggest this equation:
2 hours(fraction of work by Jason during one hour + fraction of work by Wendy during one hour)=1 work

2(1/x + 1/2x)=1

least common multiple=2x
2(2+1)=2x
2(3)=2x
6=2x
x=6/2
x=3

answer: 3 hours would ti take Jason to wash the van by himself.


5 0
3 years ago
Jake's final grade in his science class is calculated by finding the mean of his scores for six project reports. The score Jake
Yanka [14]
Let lowest possible score be x (66+ 80+ 88+ 82 +72+ x)/6=80 on solving we get ans as 92
6 0
3 years ago
Given JM=27, ml=16, jl=46, nk=15, KLM= 48, JkM=78, MJL=22, find each missing value.
Llana [10]

Answer:

KL = 27

JK = 16

MK = 30

NL = 23

m∠JKL = 132°

m∠KLJ = 22°

m∠KMJ = 54°

m∠KJL = 26°

Step-by-step explanation:

The given parameters of the quadrilateral JKLM are;

JM = 27, ML = 16, JL = 46, NK = 15, KLM = 48, JKM = 78, MJL = 22

Taking the sides as parallel, we have that quadrilateral JKLM is a parallelogram

Therefore;

KL = JM = 27

JK = ML = 16

m∠KLJ = m∠MJL = 22°

MN = NK = 15

MK = MN + NK = 15 + 15 = 30

NL = JL/2 = 46/2 = 23

m∠KJM = m∠KLM = 48°

m∠KJL = m∠KLM - m∠MJL = 48° - 22° = 26°

m∠KML = m∠JKM = 78°

m∠MKL = 180° - m∠KML - m∠KLM = 180° - 78° - 48° = 54°

m∠MKL = 54°

m∠JKL = m∠JKM + m∠MKL = 78° + 54° = 132°

m∠KMJ = m∠MKL = 54°

3 0
2 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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