1 year ago

Triangle cde maps to triangle lmn with the transformation (x,y) —> (x+3,y-2) —> (2/3x, 2/3y)

Mathematics

1 answer:

Mamont248 [21]1 year ago

4 0

The answer is 890 best answer

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In previous months, you ordered 80 boxes of pens per month. This month, you want to decrease the number of boxes of pens you ord

Lapatulllka [165]

80(0.40)=32
80-32=48 boxes of pens

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1 year ago

Read 2 more answers

La distancia de Urano al Sol es 2 870 990 000 km y la distancia de la Tierra al Sol es 1,496 × 108 km. Aproximadamente, ¿cuántas

Leni [432]

Answer:

2,7204x 10^9

Step-by-step explanati

2870000000-

149600000

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2 years ago

15,000 blue trout were released into the Meherrin River for a scientific study. The function, f(x) = 15,000 (98)x , represents t

Norma-Jean [14]

Answer:

I'm guessing you mean f(x)=15,000(9/8)^x. If this is what you mean, the population would increase by about 12,000 (12030.4870605 to be exact).

Step-by-step explanation:

Starting equation: f(x)=15,000(9/8)^x

You can clean up the 9/8 to be 1.125

Now what you want to do is find the answer to (9/8)^5 which is 1.8020324707

Next multiply 1.8020324707 by 15,000 and you get 27030.4870605

Finally 27,030.4870605 - 15,000 gives you 12030.4870605. Which means that the population increased by about 12,000.

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2 years ago

The summer camp Jessica attends is four hundred twenty-three and four tenths miles from her home. A detour adds 10 miles to the

juin [17]

Given that Jessica attends summer camp at a distance of four hundred twenty-three and four tenth mile = $423\frac{4}{10} = 423 + \frac{4}{10} = 423+0.4 = 423.4 miles$

And a detour adds 10 miles to the distance.

That means we need to add 10 miles to the given distance.

So new distance = 423.4 + 10 = 433.4 miles

Hence Jessica needs to travel 433.4 miles for summer camp.

5 0

1 year ago

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the

solniwko [45]

Answer:

(5953.52,6046.49)

Step-by-step explanation:

We are given the following in the question:

Mean, $\mu$ = 6,000 pounds

Sample size, n = 40

Alpha, α = 0.05

Standard deviation, σ = 150 pounds

95% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$6000 \pm 1.96(\dfrac{150}{\sqrt{40}} )\\\\ = 6000 \pm 46.4854=\\(5953.5146,6046.4854)\approx (5953.52,6046.49)$

are the limits within which 95% of the sample means occur.

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1 year ago