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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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The mean high temperature on a particular day in January is 31 degrees F and the standard deviation is 8.7 degrees. One year, th
geniusboy [140]

Answer:

subject?

Step-by-step explanation:

5 0
2 years ago
(-0.25x -8)-(5.5x + 9.4) the simplified expression
Alja [10]

Answer:

=−5.75x−17.4

Step-by-step explanation:

=−0.25x−8+−1(5.5x+9.4)

=−0.25x+−8+−1(5.5x)+(−1)(9.4)

=−0.25x+−8+−5.5x+−9.4

another way >>>><<<<

=−0.25x+−8+−5.5x+−9.4

=(−0.25x+−5.5x)+(−8+−9.4)

=−5.75x+−17.4

Answer:

=−5.75x−17.4

3 0
3 years ago
I need help with number 18
algol13

Answer:

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Step-by-step explanation:

4 0
3 years ago
In the distribution, the first quartile,median and mean are 30.8,48.5 and 42.0 respectively. If the co efficient skewness is -0.
finlep [7]

Answer:

The third quartile is 56.45

Step-by-step explanation:

The given parameters are;

The first quartile, Q₁ = 30.8

The median or second quartile, Q₂ = 48.5

The mean, \bar x = 42.0

Coefficient of skewness = -0.38

The Bowley's coefficient of skewness (SK) is given as follows;

SK = \dfrac{Q_3 + Q_1 - 2 \times Q_2}{Q_3 - Q_1}

Plugging in the values, we have;

-0.38 = \dfrac{Q_3 + 30.8 - 2 \times 48.5}{Q_3 - 30.8}

Which gives;

-0.38×(Q₃ - 30.8) = Q₃ + 30.8 - 2 × 48.5

11.704 - 0.38·Q₃ = Q₃ - 66.2

1.38·Q₃ = 11.704 + 66.2 =  77.904

Q₃ = 56.45

The third quartile = 56.45.

5 0
3 years ago
Simplify.<br> 3/3-n - n/n-3
Black_prince [1.1K]
\frac{3}{3-n}- \frac{n}{n-3}= \frac{3}{3-n}-(- \frac{n}{3-n})= \frac{3}{3-n}+\frac{n}{3-n}= \frac{3+n}{3-n}
8 0
3 years ago
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