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2 years ago

John is a quarterback. This year, he completed 350 passes, which is 70 percent of all the passes he's attempted this year.

Mathematics

1 answer:

Ahat [919]2 years ago

7 0

So we're given the information that 350 is 70% of "x".

So let's make "x" the total amount of attempted passes.

We can re-write 70% like this: 70/100.

This means that 70/100 * x = 350.

Solve:

70/100x = 350.

Multiply 10 on both sides to get normal number:

70/100x * 10 = 350 * 10.

700/100x = 3500

7x = 3500

Divide by 7:

7x/7 = 3500/7

x = 500

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2 years ago

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.0 and 56.

barxatty [35]

Answer: 0.05

Step-by-step explanation:

Given : Interval for uniform distribution : [46.0 minutes, 56.0 minutes]

The probability density function will be :-

$f(x)=\dfrac{1}{56-46}=\dfrac{1}{10}=0.1\ \ , 46$

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-

$P(50.75$

Hence, the probability that a given class period runs between 50.75 and 51.25 minutes = 0.05

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2 years ago

A box in a supply room contains 18 compact fluorescent lightbulbs, of which 6 are rated 13-watt, 8 are rated 18-watt, and 4 are

V125BC [204]

Answer:a) Probability P(exactly 2 bulbs rated 13watts)= 0.22

b) Probability(each bulb different rating)

= 0.24

Step-by-step explanation:

There are 6 13watts bulbs

8 18watts bulbs

4 23watts bulbs

Total bulbs = 18

a) Probability that all 3 bulbs are 18watts

Number of ways of pulling 3 bulbs = 18!/(3!×15!) = (6.4×10^15)/7.846×10^12) = 815 ways

Different ways of pulling 13watts bulbs out of 6 = 6!/(2!×4!)= 720/48=15ways

Different ways of pulling non 13 watts bulbs= 12!/(1!×11!) = 479,001,600/ 39,916,800 = 12

Number of ways total= 15×12=180waya

Therefore P(exactly 2 bulbs rated 13watts)= 180/815 =0.22

b) Probability P( all 3 bulbs are 1 from each rating)

Ways of pulling 3bulbs bulb each from 3 ratings are 4× 5 × 8= 192ways

Probability = 192/815 =0.24

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2 years ago

Read 2 more answers

I needdd helppp fast plzzz

Alexandra [31]

Answer:

Measure of ∠B = 51°.

Step-by-step explanation:

The sum of the measures of any triangle will always add up to 180°. Also the sum of two supplementary angles, or angles that are side by side on a straight line, will also add up to 180°. In order to find the measure of ∠B, you need to first find the measure of ∠C. ∠C is supplementary to the angle of 113°, so we can find ∠C: 180 - 113 = 57°. Now that we know ∠A and ∠C, we can solve for ∠B: 180 = 72 + 57 + ∠B or 180 = 129 + ∠B, subtracting 129 from both sides, we get ∠B = 51°.

4 0

3 years ago

The probability of passing the math class of Professor Goodrum is 59%, the probability of passing Professor Cruise's physics

liq [111]

Answer:

51%

Step-by-step explanation:

Given P(passing math) = 59%, P(passing physics) = 26%, and P(passing both) = 17%, you want to find the probability of passing only one of the courses.

<h3>Probability relations</h3>

We can record the given probabilities in a 2-way table (values shown in blue). The table is completed by making sure the totals add up (values shown in black).

The probability of passing one course and failing the other is the sum of the probabilities with a yellow background:

42% +9% = 51%

The probability of passing one or the other is 51%.

<em>Additional comment</em>

We can also get there using the relation ...

P(A+B) = P(A) +P(B) -P(AB)

The union of A and B also includes their overlap:

P(A+B) = P(AB') +P(A'B) +P(AB)

In other words, the probability of interest is ...

P(AB') +P(A'B) = P(A) +P(B) -2×P(AB) = 59% +26% -2(17%)

P(AB') +P(A'B) = 51%

6 0

9 months ago