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3 years ago

Find the percent decrease. Round to the nearest percent. From 96°F to 71°F

Mathematics

1 answer:

barxatty [35]3 years ago

7 0

Step-by-step explanation:

100% = 96

1% = 100%/100 = 96/100 = 0.96

the temperature changed by 96 - 71 = 25.

how many % are these 25 ?

we check how often 1% fits into 25.

that is done by division :

25 / 0.96 = 26.04166667 ≈ 26%

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larisa86 [58]

Answer:

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mrs_skeptik [129]

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in the figure below trianglePQM and triangleQRP are right triangles. the measure of lineQM is 6 and the measure of lineQP is 8.

Kipish [7]

Answer:

Option 4 is correct. The length of PR is 6.4 units.

Step-by-step explanation:

From the given figure it is noticed that the triangle PQR and triangle MQR.

Let the length of PR be x.

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$hypotenuse^2=base^2+perpendicular^2$

Use pythagoras formula for triangle PQM.

$PM^2=QM^2+PQ^2$

$PM^2=(6)^2+(8)^2$

$PM^2=36+64$

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The value of PM is 10. The length of PR is x, so the length of MR is (10-x).

Use pythagoras formula for triangle PQR.

$PQ^2=QR^2+PR^2$

$(8)^2=QR^2+x^2$

$64-x^2=QR^2$ .....(1)

Use pythagoras formula for triangle MQR.

$MQ^2=QR^2+MR^2$

$(6)^2=QR^2+(10-x)^2$

$36=QR^2+x^2-20x+100$

$36-x^2+20x-100=QR^2$ .... (2)

From equation (1) and (2) we get

$36-x^2+20x-100=64-x^2$

$20x-64=64$

$20x=128$

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3 years ago

Solvex + 5-6 = 7.

yKpoI14uk [10]

I got x=8
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2 years ago

Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (

Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

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Given that

P(S) = 0.3

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$P(\frac{T}{S^c}) = 0.05$

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

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= 1 - 0.3

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ii. $P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}$

$= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }$

$= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}$

= 0.8906

iii. $P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}$

$= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }$

$= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}$

= 0.0221

We simply applied the above formulas so that the each part could come

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3 years ago

Find the percent decrease. Round to the nearest percent. From 96°F to 71°F​

Find the percent decrease. Round to the nearest percent. From 96°F to 71°F