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3 years ago

If you know two figures are congruent, then what would be true about all of the corresponding parts of the two figures?

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

Answer:

if their corresponding sides are equal in length, and their corresponding angles are equal in measure.

Step-by-step explanation:

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1,200 students registered for an art camp. 945 students showed up. calculate the precent error

Reil [10]

Answer:

255

Step-by-step explanation:

1200-945 is 255

7 0

4 years ago

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MA = 8x – 2, mB = 2x – 8, and mC = 94 – 4x. List the sides of ABC in order from shortest to longest.

lutik1710 [3]

I hope this helps you

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3 years ago

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Triangle ABC, with vertices A(5,2), B(7,6), and C(2,5), is drawn inside a rectangle, as shown below.

natka813 [3]

Answer:

9 units²

Step-by-step explanation:

calculate the area of Δ ABC as

area of rectangle - area of 3 outer right triangles

area of rectangle = 5 × 4 = 20 units²

area of left Δ = $\frac{1}{2}$ × 3 × 3 = 4 $\frac{1}{2}$ units²

area of Δ on right = $\frac{1}{2}$ × 2 × 4 = 4 units²

area of top Δ = $\frac{1}{2}$ × 5 × 1 = 2 $\frac{1}{2}$ units²

Then

area of Δ ABC = 20 - 4 $\frac{1}{2}$ - 4 - 2 $\frac{1}{2}$ = 20 - 11 = 9 units²

8 0

1 year ago

The circumference of a circle is 56.52 feet. What is the circle's radius?

matrenka [14]

Answer:

If we substitute pi as 3.14 we would be doing -

56.52/3.14 = 18

18/2 = 9.

So the circle's radius would be 9.

BUT if you want it with PI. . .

The answer would be 8.995437385

7 0

3 years ago

(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is $P_T= 0.4560$

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$

$=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ]$

$=0.001925 -0.000625$

$P_F= 0.0013$

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3 years ago

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