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3 years ago

How do I find the area???

Mathematics

1 answer:

V125BC [204]3 years ago

4 0

Answer:

option A. 36 square inches

Step-by-step explanation:

The figure is a trapezium

Area of a trapezium=1/2 *(sum of parallel sides) *height

From the diagram, the parell sides are 6 inches and 12 inches ,and the height is 4 inches.

We substitute these values into the formula.

This implies that, the area

$= \frac{1}{2}(6 + 12) \times 4$

$= \frac{1}{2}(18) \times 4$

$=9\times 4$

$= 36$

Thus the area is 36 square inches

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8. Temperatures in degrees Celsius (C) may be changed to degrees Fahrenheit (F) by using the formula F = 1.8C + 32º. Find the te

pentagon [3]

Answer:

-40 °F = -40 °C

Step-by-step explanation:

(1) F = 1.8 C + 32

If the Fahrenheit and Celsius temperatures are the same, then

(2) F = C Substitute (2) into (1)

C = 1.8C + 32 Subtract 1.8 C from each side

C – 1.8 C = 32 Combine like terms

-0.8C = 32 Divide each side by -0.8

C = -32/0.8

(3) C = -40 Substitute (3) into (1)

F = 1.8(-40) + 32

F = -72 + 32

F = -40

So, -40 °F = -40 °C

The thermometer below shows that -40 °F = -40 °C.

8 0

3 years ago

Kayleigh wants to buy a car that costs $13,080.00. She deposits $12,000.00 into a savings account that earns 3% simple interest.

Ipatiy [6.2K]

The formula is
A=p (1+rt)
A cost of the car 13080
P amount deposited 12000
R interest rate 0.03
T time?
Solve for t
T=[(A/p)-1]÷r
T=((13,080÷12,000)−1)÷0.03
T=3 years

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3 years ago

Multiplying and Dividing Rational Numbers

Dafna11 [192]

Answer:

the answer is 9.9

Step-by-step explanation:

4 0

3 years ago

The vertex of the parabola is at (-3,-2) which of the following could be its equation

ki77a [65]

D;x=-2(y+2)^2-3 is the equation

3 0

2 years ago

Read 2 more answers

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

3 0

3 years ago