If you pick a card at random from a standard 52-card deck, what is the probability that it is either red or a 7?

Answers

Answer 1

Answer: The probability that the card you pick from a standard 52-card deck is 7/13.

Let's first solve how many cards out of the 52 are red.
Since half the deck is red and the other half is black, 1/2 of 52 is 26 cards.

Next, let's solve how many cards you can choose with seven on it.
Since there are four of each kind of card in a standard deck of cards, there are four sevens. However, we already counted the two red sevens when we solved how many red cards there are. So 4-2 = 2 cards.

26 + 2 = 28 cards
28/52 = 7/13.

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A cone is inside a cylinder. The cone and the cylinder have the same base and height.The part of the cylinder that is NOT taken up by the cone has twice the volume of the cone. What fraction of the total cylinder is taken up by the cone?A. 3/2B. 2/3C. 1/2D. 1/3

IQ scores are normally distributed with a mean of 105 and a standard deviation of 17. Assume that many samples of size n are taken from a large population of people and the mean IQ score is computed for each sample. If the sample size is n 81, find the mean and standard deviation of the distribution of sample means.

Answers

Answer:

Mean 105

Standard deviation 1.89

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = (\sigma)/(√(n))$

In this problem, we have that:

$\mu = 105, \sigma = 17$

If the sample size is n 81, find the mean and standard deviation of the distribution of sample means.

By the Central limit theorem

mean 105

Standard deviation

$s = (17)/(√(81)) = 1.89$

I do have a time limit. I appreciate any helpIf YB = ZA find the value of X and the length of YB

Answers

Since ZA and YB are equal in length and the two lines are parallel to each other.

Also, we can see that YZ and AB are perpendicular to both ZA and YB, thus

YZ = AB

16x - 4 = 4 - 4x

16x + 4x = 4 + 4

20x = 8

x= 8/20 = 2/5

YB = ZA = 20x - 5 = 20(2/5) - 5 = 3

Thus, the value of x is 2/5 and length of YB is 3 units.

So, the correct answer is option A

The owner of a fabric store has determined that the profits P of the store are approximately given by P(x) = -x^2 + 70x+67, where x is the yards of fabric sold daily. Find the maximum profit to the nearest dollar. a) $617 b) $792 c) $1017 d) $1292 e) none

Answers

Answer: Option 'd' is correct.

Step-by-step explanation:

Since we have given that

Profit function of the store is given by

$P(x)=-x^2+70x+67$

We need to find the maximum profit.

For this, we first derivate the above function:

$P'(x)=-2x+70$

Now, put P(x) = 0, we get that

$-2x+70=0\n\n-2x=-70\n\nx=35$

Now, we will check that its maximality by finding the second derivative:

$P''(x)=-2<0$

it gives maximum profit at x = 35 yards.

And the maximum profit would be

$P(35)=-(35)^2+70* 35+67=\$1292$

Hence, Option 'd' is correct.

Find sin (A-B) if sin A = 4/5 with A between 90 and 180 and if cos B = 3/5 with B between 0 and 90

Answers

Answer:

sin(A-B) = 24/25

Step-by-step explanation:

The trig identity for the differnce of angles tells you ...

sin(A -B) = sin(A)cos(B) -sin(B)cos(A)

We are given that sin(A) = 4/5 in quadrant II, so cos(A) = -√(1-(4/5)^2) = -3/5.

And we are given that cos(B) = 3/5 in quadrant I, so sin(B) = 4/5.

Then ...

sin(A-B) = (4/5)(3/5) -(4/5)(-3/5) = 12/25 + 12/25 = 24/25

The desired sine is 24/25.

Before 1918, approximately 40% of the wolves in a region were male, and 60% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 60% of wolves in the region are male, and 40% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (Round your answers to three decimal places.) (a) Before 1918, in a random sample of 10 wolves spotted in the region, what is the probability that 7 or more were male

Answers

Answer:

P(≥ 7 males) = 0.0548

Step-by-step explanation:

This is a binomial probability distribution problem.

We are told that Before 1918;

P(male) = 40% = 0.4

P(female) = 60% = 0.6

n = 10

Thus;probability that 7 or more were male is;

P(≥ 7 males) = P(7) + P(8) + P(9) + P(10)

Now, binomial probability formula is;

P(x) = [n!/((n - x)! × x!)] × p^(x) × q^(n - x)

Now, p = 0.4 and q = 0.6.

Also, n = 10

Thus;

P(7) = [10!/((10 - 7)! × 7!)] × 0.4^(7) × 0.6^(10 - 7)

P(7) = 0.0425

P(8) = [10!/((10 - 8)! × 8!)] × 0.4^(8) × 0.6^(10 - 8)

P(8) = 0.0106

P(9) = [10!/((10 - 9)! × 9!)] × 0.4^(9) × 0.6^(10 - 9)

P(9) = 0.0016

P(10) = [10!/((10 - 10)! × 10!)] × 0.4^(10) × 0.6^(10 - 10)

P(10) = 0.0001

Thus;

P(≥ 7 males) = 0.0425 + 0.0106 + 0.0016 + 0.0001 = 0.0548

70 = 10 + 12r solve each equation

Answers

Answer:

r = 5

Step-by-step explanation:

70 = 10 + 12 r

-10 -10

____________

60 = 12r

/12 /12

_______

5 = r

Answer:

Step-by-step explanation:

70-10=60

60=12r

60/12=5

r=5

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