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3 years ago

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just olya [345]3 years ago

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Solve for h<br><br> can someone please answer and explain thank you <33

Oxana [17]

Answer: $h=\frac{2A}{(a+b)}$

Step-by-step explanation:

To solve for h, we want to get h alone. We need to use our algebraic properties.

$A=\frac{1}{2} h(a+b)$ [multiply both sides by 2]

$2A=h(a+b)$ [divide both sides by (a+b)]

$h=\frac{2A}{(a+b)}$

Now that we have h alone, we know that $h=\frac{2A}{(a+b)}$ .

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3 years ago

Simplify equation: 6x2+7x−3x2−2x

tigry1 [53]

Answer:

8x+4

Step-by-step explanation:

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3 years ago

At an elevation of 1221.4 feet, Lake Mead will hold 28,945,000 acre-feet of water. Which number is the best estimate of this amo

uranmaximum [27]

Answer:

$3 \times 10^{7}$

Step-by-step explanation:

At an elevation of 1221.4 feet, Lake Mead will hold 28,945,000 acre-feet of water.

So, the amount of water that is in the lake will be 28,945,000 acre-feet which is estimated to be 30000000 acre-feet of water.

Therefore, the number $3 \times 10^{7}$ is the best estimate of this amount of water. (Answer)

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3 years ago

A water tank currently contains 275 gallons of water. The amount of water in the tank will decrease at a constant rate of 15 gal

makvit [3.9K]

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0

kolbaska11 [484]

Answer:

$Q_3 = 56.45$ --- The third quartile

$Var = 370.18$ -- Variance

Step-by-step explanation:

Given

$Q_1 = 30.8$ -- First quartile

$Q_2 = 48.5$ --- Median

$\bar x = 42$ --- Mean

$Skp = -0.38$ --- Coefficient of skewness

Solving (9): The third quartile $Q_3$

This is calculated from

$Skp = \frac{Q_1 + Q_3 - 2Q_2}{Q_3 - Q_1}$

So, we have:

$-0.38 = \frac{30.8 + Q_3- 2*48.5}{Q_3 - 30.8}$

Cross Multiply

$-0.38 (Q_3 - 30.8)= 30.8 + Q_3- 2*48.5$

Open bracket

$-0.38Q_3 + 11.704= 30.8 + Q_3- 97.0$

Collect like terms

$-0.38Q_3 -Q_3= 30.8 - 97.0- 11.704$

$-1.38Q_3= -77.904$

Divide both sides -1.38

$Q_3 = \frac{-77.904}{-1.38}$

$Q_3 = 56.45$ --- approximated

Solving (b): The variance

First, calculate the standard deviation from:

$3IQR = 4SD$

$IQR= Q_3 - Q_1$

So:

$3IQR = 4SD$

$3(Q_3 - Q_1) = 4SD$

Make SD the subject

$SD = \frac{3}{4}(Q_3 - Q_1)$

$SD = \frac{3}{4}(56.45 - 30.8)$

$SD = \frac{3}{4}*25.65$

$SD = \frac{3*25.65}{4}$

$SD = \frac{76.95}{4}$

$SD = 19.24$

So, the variance is:

$Var = SD^2$

$Var = 19.24^2$

$Var = 370.18$

5 0

3 years ago