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andreev551 [17]
2 years ago
7

Please help and thanks

Mathematics
1 answer:
zhenek [66]2 years ago
6 0

a right angle so 90 degrees

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Use synthetic division to perform (x^4 - 3x + 5) = (x - 4).
Anastaziya [24]

Answer:

C is correct

Step-by-step explanation:

4 0
3 years ago
A circle of yellow tulips is planted in Cedarburg's central park. Pam measured the circle and calculated that is has a circumfer
lozanna [386]

Answer:

The circle's diameter is 4\ yd

Step-by-step explanation:

we know that

The circumference of a circle is equal to

C=\pi D

where

D is the diameter of the circle

In this problem we have

C=12.56\ yd

\pi=3.14

substitute the values and solve for D

12.56=(3.14)D

D=12.56/(3.14)=4\ yd

6 0
3 years ago
One alloy is 2 parts iron to 3 parts silver and another alloy is 7 parts iron to 3 parts silver. How much of each should be comb
Shalnov [3]
Let x be your first alloy
2/5x+7/10(30-x)=1/2(30)
4x+210-7x=150
3x=60
x=20
So, you need 20 lbs of the first alloy, and 10 parts of the second to make 30 lbs of half iron and half silver alloy!
8 0
3 years ago
A violin student records the number of hours she spends practicing during each of nine consecutive weeks: 6.2 5.0 4.3 7.4 5.8 7.
spin [16.1K]

Answer:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

Step-by-step explanation:

Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:

6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3

For this case we need to sort the data first on increasing way and we got:

1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4

For this case we have 9 values the median would be on the 5 position:

Median = 6.2

The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0

The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2

Then the interquartile rnage would be:

IQR = Q_3 -Q_1= 7.2-5=2.2

And 1.5 IQR = 1.5*2.2=3.3

The lower limit on this case would be:

LL= Q_1 -1.5 IQR= 5-3.3=1.7

And our value is 1.2 is lower than the lower limit 1.7

Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below  the first quartile.

False 1.2 is more than 1.5 IQR below the first quartile

C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  median.

False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median

D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  mean.

False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean

5 0
3 years ago
What is the scale factor if the scale is 10 inches = 1 foot?
klasskru [66]

10 inches to 12 inches

10:12

5:6

<em>Have a Great Day</em>

5 0
3 years ago
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