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3 years ago

Find the area of a trapezoid with bases that measure 9.2 CM and 7.7 cm and height 6.8 cm

2 answers:

8 0

$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\
---------\\
a=9.2\\
b=7.7\\
h=6.8
\end{cases}\implies A=\cfrac{6.8(9.2+7.7)}{2}$

8_murik_8 [283]3 years ago

5 0

Answer:

The area of a trapezoid is 57.46 cm².

Step-by-step explanation:

Formula

$Area\ of\ trapezoid = \frac{(a + b)}{2}\times h$

Where a, b are the two parallel bases of the trapezoid.

h is the height of the trapezoid.

As given

Trapezoid with bases that measure 9.2 cm and 7.7 cm and height 6.8 cm.

Put in the formula

$Area\ of\ trapezoid = \frac{(9.2 + 7.7)}{2}\times 6.8$

$Area\ of\ trapezoid = \frac{16.9}{2}\times 6.8$

$Area\ of\ trapezoid = 8.45\times 6.8$

Area of trapezoid = 57.46 cm²

Therefore the area of a trapezoid is 57.46 cm².

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x = -2 represents a vertical line that intersects the x axis at the point (-2, 0). Since the line is vertical, the slope is undefined making it not have a positive slope.

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3 years ago

Need help with #9 please

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3 years ago

For what value of x is the rational expression below undefined x-4/3x2-75

abruzzese [7]

The given expression is undefined when x = 5 OR x = -5

<h3>Undefined expression </h3>

From the question, we are to determine the value of x for which the given expression is undefined

The given expression is

x-4/3x²-75

An expression is undefined when the denominator equals zero

Thus, the expression is undefined when

3x² - 75 = 0

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2 years ago

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = the number that are not correctly calibrated.

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = 0.35

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = 0.54

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = 0.11

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

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3 years ago