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3 years ago

How do you use an exponent to represent a number such as 16

Mathematics

1 answer:

lora16 [44]3 years ago

5 0

This can be a perfect square! which 4x4
that is 4^2

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Which number is a rational number?

motikmotik

17.156 is rational. Hope that helped.

3 0

2 years ago

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume

kobusy [5.1K]

Answer:

The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018

Step-by-step explanation:

The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

$W = \frac{X - \mu}{\sigma} = \frac{X - 182}{6.5465}$

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.

The cummulative distribution function of W will be denoted by $\phi$ . The values of $\phi$ can be found in the attached file.

$P(X > 201.0476) = P(\frac{X-182}{6.5465} > \frac{201.0476 - 182}{6.5465}) = P(W > 2.91) = 1-\phi(2.91) = \\1-0.9982 = 0.0018$

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.

Download pdf

6 0

4 years ago

Please help me find the area of this figure.

GREYUIT [131]

You basically have a triangle and a rectangle. seperate the two at first, the rectangle's area is 32 since 8*4=32
the area of a triangle is half of it's base times width. so 3*5=15/2=7.5
add 7.5 and 32 to get 39.5

5 0

3 years ago

If ax-7=b(x+2) then x =

Maslowich

Move all terms to the left side and set equal to zero. Then set each factor to equal to zero giving you..

8 0

3 years ago

Identify all pairs of congruent corresponding parts. Then write another congruence statement for the triangles.

astra-53 [7]

SOLUTION

$\begin{gathered} \Delta FED\cong\Delta CBA \\ \angle F\cong\angle C \end{gathered}$ $\begin{gathered} \Delta EFD\cong BCA \\ \angle E\cong\angle B \\ \bar{DE}\cong\bar{AB} \end{gathered}$ $\begin{gathered} \bar{EF}=\bar{BC} \\ \angle D=\angle A \\ \bar{DF}=\bar{AC} \end{gathered}$

4 0

2 years ago