Solve -2x + 13 = -7x + 28

telo118 [61]

I don't understand your question are you multiplying or what

7 0

3 years ago

The distribution of the amount of money spent by students on textbooks in a semester is approximately normal in shape with a mea

zheka24 [161]

Answer:

$301 - $397

Step-by-step explanation:

Using the Empirical rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2)95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ .

From the above question,

Mean = 349 , standard deviation = 24.

Confidence interval = 95%

Using 2)95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

μ – 2σ

= 349 - 2(24)

= 349 - 48

= 301

μ + 2σ

349 + 2(24)

= 349 + 48

= 397

Therefore, according to the standard deviation rule, approximately 95% of the students spent between $301 and $397 on textbooks in a semester.

8 0

3 years ago

When the members of a family discussed where their annual reunion should take place, they found that out of all the family mem

Doss [256]

Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those that would only not go to a park, $A \cap B$ are those who would not got to a park or to the beach, $A \cap C$ are those who would not go to a park or to the famili cottage. And $A \cap B \cap C$ are those that would not go to any of these places.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following values:

$a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The total number of family members is the sum of all these values:

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family cottage.

This means that $A \cap B \cap C = 5$

1 would go to all three places. This means that $d = 1$ .

8 would go to neither a park nor the family cottage

This means that:

$A \cap C + (A \cap B \cap C) = 8$

$A \cap C = 3$

8 would go to neither a beach nor the family cottage

$B \cap C + (A \cap B \cap C) = 8$

$B \cap C = 3$

7 would go to neither a park nor a beach

$A \cap B + (A \cap B \cap C) = 7$

$A \cap B = 2$

15 would not go to the family cottage

$C = 15$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$15 = c + 3 + 3 + 5$

$c = 4$

12 would not go to a beach

$B = 12$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$12 = b + 3 + 2 + 5$

$b = 2$

11 would not go to a park

$A = 11$

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$11 = a + 2 + 3 + 5$

$a = 1$

Now, we can find the total number of family members.

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5$

$T = 21$

The total number of family members is 21.

5 0

3 years ago

T(x) = ax^5 + 2 where is a real number. If the points (-2, 66) and (2, -62) are on the graph of the function classify as one of

Sergeu [11.5K]

Answer:

a < -1

Step-by-step explanation:

Okay, so we are given this equation:

t(x) = ax^5 + 2

Lets plug in -2 as x and see what happens.. t(x) can be 66..

$a * (-2)^{5} + 2$ = 66

-32a + 2 = 66

-32a = 64

a = -2

A is less than -1,

a < -1

6 0

3 years ago

If m=8, how many solutions are there for the expression m-8?

ss7ja [257]

Answer:

1 solution, 0

Step-by-step explanation:

In algebra, a variable can be equal to any amount of numbers depending on how it is used. In this case, m is only equal to the one value of 8, and therefore only has <u>one solution.</u>

m - 8

(8) - 8

0

5 0

3 years ago