Answer:
-6 + 3x = -9

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And 6 to both sides:

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3x = -3

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Divide by 3 on both sides:

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x = -1

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**Answer: x = -1**

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And 6 to both sides:

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3x = -3

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Divide by 3 on both sides:

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x = -1

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One cell phone company offers a plan that costs $25.99 and includes unlimited night and weekendminutes. Another company offers a plan that costs $14.99 and charges $0.35 per minute during nightsand weekends. For what numbers of night and weekend minutes does the second company's plan costmore than the first company's plan?

Simplify \sqrt{25a^2}

Helpppppp , quick answer with work pleaseeee

1. number 5 added to three times the product of M and N.2. product of number Y and Z subtracted from 10.3. sum of number X and Y subtracted from their products.4. workings pls

I thought of a number. This number double is 7 less than 29. What is my number? hurry please

Simplify \sqrt{25a^2}

Helpppppp , quick answer with work pleaseeee

1. number 5 added to three times the product of M and N.2. product of number Y and Z subtracted from 10.3. sum of number X and Y subtracted from their products.4. workings pls

I thought of a number. This number double is 7 less than 29. What is my number? hurry please

**Answer:**

a) The cumulative distribution function would be given by:

x 0 1 2 3 4 5

F(X) 0.05 0.15 0.30 0.55 0.9 1

b)

And replacing we got:

**Step-by-step explanation:**

For this case we have the following probability distribution function given:

x 0 1 2 3 4 5

P(X) 0.05 0.1 0.15 0.25 0.35 0.1

We satisfy the conditions in order to have a probability distribution:

1)

2)

**Part a**

The cumulative distribution function would be given by:

x 0 1 2 3 4 5

F(X) 0.05 0.15 0.30 0.55 0.9 1

**Part b**

For this case we want to find this probability:

And replacing we got:

b. What is the probability that at least one tank in the sample contains high-viscosity material?

c. In addition to the six tanks with high viscosity levels, four different tanks contain material with high impurities. What is the probability that exactly one tank in the sample contains high-viscosity material and exactly one tank in the sample contains material with high impurities?

**Answer:**

**a) P(A) = 0,4607 or P(A) = 46,07 %**

**b) P(B) = 0,7120 or 71,2 %**

**c) P(C) = 0,2055 or P(C) = 20,55 %**

**Step-by-step explanation:**

**We will use two concepts in solving this problem.**

**1.- The probability of an event (A) is for definition:**

**P(A) = Number of favorable events/ Total number of events FE/TE**

**2.- If A and B are complementary events ( the sum of them is equal to 1) then:**

**P(A) = 1 - P(B)**

**a) The total number of events is:**

**C ( 24,4) = 24! / 4! ( 24 - 4 )! ⇒ C ( 24,4) = 24! / 4! * 20!**

**C ( 24,4) = 24*23*22*21*20! / 4! * 20! **

**C ( 24,4) = 24*23*22*21/4*3*2**

**C ( 24,4) = 24*23*22*21/4*3*2 ⇒ C ( 24,4) = 10626**

**TE = 10626**

**Splitting the group of tanks in two 6 with h-v and 24-6 (18) without h-v**

**we get that total number of favorable events is the product of:**

**FE = 6* C ( 18, 3) = 6 * 18! / 3!*15! = 18*17*16*15!/15!**

**FE = 4896**

**Then P(A) ( 1 tank in the sample contains h-v material is:**

**P(A) = 4896/10626**

**P(A) = 0,4607 or P(A) = 46,07 %**

**b) P(B) will be the probability of at least 1 tank contains h-v **

**P(B) = 1 - P ( no one tank with h-v)**

**Again Total number of events is 10626**

**The total number of favorable events for the ocurrence of P is C (18,4)**

**FE = C (18,4) = 18! / 14!*4! = 18*17*16*15*14!/14!*4!**

**FE = 18*17*16*15/4*3*2 = 3060**

**Then P = 3060/10626**

**P = 0,2879**

**And the probability we are looking for is**

**P(B) = 1 - 0,2879**

**P(B) = 0,7120 or 71,2 %**

**c) We call P(C) the probability of finding exactly 1 tank with h-v and t-i**

**having 4 with t-i tanks is:**

reasoning the same way but now having 4 with t-i (impurities) number of favorable events is:

FE = 6*4* C(14,2) = 24 * 14!/12!*2!

FE = 24* 14*13*12! / 12!*2

FE = 24*14*13/2 ⇒ FE = 2184

**And again as the TE = 10626**

**P(C) = 2184/ 10626**

**P(C) = 0,2055 or P(C) = 20,55 %**

These problems relate to the** field of probability** and specifically utilize the Hypergeometric Distribution. By plugging data into the appropriate formula, we can find the probabilities. For instance, the joint probability of independent events can be used to find the chance of exactly one tank having high viscosity and one tank having high impurities.

This question is asking us to solve probability problems. It specifically related to the **Hypergeometric Distribution**, which is used when we're interested in success/failure outcomes (in this case, tanks with high or acceptable viscosity), and when we're sampling without replacement.

For part a, we are looking for the probability of picking exactly one tank with high-viscosity. We would calculate this using the hypergeometric distribution formula:

**P(X=k) = (C(K, k) * C(N-K, n-k)) / C(N, n)**

where K is the total number of success states in the population (6 tanks with high viscosity), k is the number of success states in the sample (1 tank), N is the population size (24 tanks), and n is the number of samples (4 tanks). Plugging these numbers in, we can find the answer.

For part b, to find the probability that at least one tank in the sample contains **high-viscosity material**, we can either sum the probabilities P(X=1), P(X=2), P(X=3), and P(X=4), or find the complement of the probability that none of the tanks have high viscosity, i.e., 1- P(X=0).

For part c, with the addition of 4 tanks with high impurities, we can use the joint probability of independent events, which is the product of the probabilities of the two independent events. Here, the probability that exactly one tank has high viscosity and exactly one tank has high impurities would be the product of the two individual probabilities calculated in a similar manner.

#SPJ3

**Answer:**

**Therefore the general solution is**

**Step-by-step explanation:**

Integration Rule:

Given differential equation is

[ multiplying dx both sides]

[ dividing both sides]

Integrating both sides

[ c is arbitrary constant]

**Therefore the general solution is**

**Answer:**

0.28

**Step-by-step explanation:**

For a certain breed of dog,

OFFSPRING OF A BLACK DOG:

Is black = 0.6 Is brown = 0.4

OFFSPRING OF A BROWN DOG:

Is black = 0.2 Is brown = 0.8

**If REX is a brown dog of this breed, what is the probability that his grand puppy will be black?**

The probability that his grand pup will be black = Probability that his black pups produce black pups + the probability that his brown pups produce black pups.

The probability that his black pups produce black pups:

0.2 × 0.6 = 0.12

The probability that his brown pups produce black pups:

0.8 × 0.2 = 0.16

The probability that his grand pup will be black = 0.12 + 0.16 = 0.28

The probability that the grand-pup of a brown dog (like Rex) will be black is calculated by multiplying the probability that a brown dog will produce a black offspring (0.2) by the probability that a black dog will produce a black offspring (0.6). This results in a probability of 0.12, or 12%.

In this problem, you're looking at the **probability** of a twice-removed descendent (a grand-pup) being a certain color, given the starting color of the ancestor (Rex). Since the color of the offspring is dependent on the color of the parent, this is a **conditional probability** problem.

The probability that the offspring of a brown dog will be black is 0.2. If that dog is black, the probability of its own offspring being black is 0.6. So, you multiply those probabilities together to find the probability that a brown dog will have a black grandpup, which is 0.2 * 0.6 = 0.12, or 12%. Therefore, the **probability** that Rex's grandpup will be black is **12%**.

#SPJ3

100, 20, 4, ...

**Answer:**

its a geometric progression. r=1/5 or 0.2

The slope of the line is 1/3

**Answer:**

1/3

**Step-by-step explanation:**

First you have to do this step to find your slope: y2-y1/x2-x1

So your first point is (0,-4) (3,-3)

I got 1/3

You can also do rise over run as well