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4 years ago

Find the outlier of the data set 3.1, 5.3, 4.3, 0.7, 5.6, 5.2, 5.8, 5.4

Mathematics

1 answer:

Vilka [71]4 years ago

5 0

0.7

Low outlier = Q1 - 1.5(IQR)
High outlier = Q3 + 1.5(IQR)

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Can you help me solve this please step by step

eimsori [14]

First, the values of 2sqrt(3) or 3sqrt(3) are not even twice the length of the given base, 3.

Next, I would say that 6 is the better answer than 12, assuming that the triangle is”drawn-to-scale”.

Finally, I think the hypotenuse is 6.

8 0

3 years ago

The midpoint of is (14, 26). One endpoint of is A(32, 40). What are the coordinates of B?

cupoosta [38]

For x,
(32+x)/2=14
32+x=14*2
32+x=28
x=28-32
x=-4

For y,
(40+y)/2=26
40+y=26*2
40+y=52
y=52-40
y=12

Therefore coordinates of B(-4,12)
Hope this helps!

7 0

3 years ago

If 3x+54+(65 × 32)+ 902- (43/12) =y, then what is y?

IceJOKER [234]

Answer:

We cannot solve without a value for x. The simplified form is $3x+\frac{36,389}{12} =y$

Step-by-step explanation:

If the equation is $3x+54+(65)(32)+902-\frac{43}{12}=y$ then we cannot solve it until we know the value of x. We can simplify it using order of operations such as PEMDAS.

$3x+54+(65)(32)+902-\frac{43}{12}=y \\3x+54+2,080+902-\frac{43}{12}=y \\3x+3,036-\frac{43}{12}=y \\3x+\frac{36,432}{12} -\frac{43}{12} =y\\3x+\frac{36,389}{12} =y$

6 0

3 years ago

Three fourths of the fruit in a refrigerator are apples. There are 24 apples in the refrigerator. How many pieces of fruit are i

Eva8 [605]

3/4=24 so 1/4=8 therefore 4/4=32

8 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

0.5+0.0833=0.5833

4 0

4 years ago