Anyone know the answer to this geometry question?

greg works at a video store he earns $7 an hour plus $5 for each memebership he sells write a formula that will help greg determ

san4es73 [151]

A= 7h+5m okayyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

8 0

3 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

2 years ago

Kyle's car repair costs $1000. He has 2 coupons which could save him $80 off of $1000, or 15% off of $1000. EXPLAIN WHICH COUPON

never [62]

Answer:

Kyle would save more using his 15% off of $1000 coupon.

Explanation:

15 ÷ 100 • 1000 = 150

1000 - 150 = $850

$150 is the amount the 15% coupon took off.

$850 is the amount Kyle will pay.

vs.

1000 - 80 = $920

6 0

2 years ago

Reina surveyed her friends, asking each of them, "How many times have you been to the amusement park?" The resulting data set ha

Anna [14]

Answer:

1. right, 2. median, 3. range or the IQR

Step-by-step explanation:

5 0

2 years ago

What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?

nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

$f(x)=x^5-10x^3+9x$ ; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

$f'(x)=5x^4-30x^2+9$

Setting derivative equal to 0, we get

$5x^4-30x^2+9=0$

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376 : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = $x^5- 10x^3+ 9x$ on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

7 0

3 years ago