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3 years ago

The trapezoid below has an area of 1575 cm. What equation could you solve to find the height of the trapezoid

1 answer:

3 0

The trapezoid formula is 1/2 (base 1+ base 2) x height. So if you work backwards you should be able to find the height. Could you give the base measurements?

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Plz help! 20 points, correct answer will be marked brainliest :)

Tanzania [10]

Answer:

D) f(x) = 6x

Step-by-step explanation:

5 0

3 years ago

Question attached plz answer

Bezzdna [24]

Answer:

x = - 2.5

Step-by-step explanation:

Given that the sketch represents

y = x² + bx + c

The graph crosses the y- axis at (0 , - 14), thus c = - 14

y = x² + bx - 14

Given the graph crosses the x- axis at (2, 0), then

0 = 2² + 2b - 14

0 = 4 + 2b - 14 = 2b - 10 ( add 10 to both sides )

10 = 2b ( divide both sides by 2 )

b = 5

y = x² + 5x - 14 ← represents the graph

let y = 0 , then

x² + 5x - 14 = 0 ← in standard form

(x + 7)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 7 = 0 ⇒ x = - 7

x - 2 = 0 ⇒ x = 2

The x- intercepts are x = - 7 and x = 2

The vertex lies on the axis of symmetry which is midway between the x- intercepts, thus

the x- coordinate of the turning point is $\frac{-7+2}{2}$ = $\frac{-5}{2}$ = - 2.5

8 0

3 years ago

Simplify 5 to the negative 4th power over 5 to the 3rd

Licemer1 [7]

"<span>5 to the negative 4th power over 5 to the 3rd</span>" translates to

5⁻⁴ / 5³

So you can just combine exponents.

5⁻⁴ / 5³ = 5⁻⁴ ⁻ ³ = 5⁻⁷ = 1/5⁷

Final answer: 1/5⁷

4 0

3 years ago

Tom Harding is a proofreader. He is paid $0.23 for every page he proofs. What is Tom's total pay for a week in which he proofrea

natka813 [3]

Alrighty, what I was always taught was to start by writing out what you know from the question.

What I know:
-Tom is paid $0.23 for every page of proofs.
-He proofreads 1,992 pages.

In order to figure out how much money Tom has made, we need to multiply 0.23 by 1,992. Multiplying allows us to see the total amount of money made by proofreading x amount of pages - for every page he proofreads, he makes $0.23.
0.23×1992=458.16

Therefore, Tom made $458.16.

Hope this helps!

3 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago