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3 years ago

2x7-10/9-4 What is the answer for this

Mathematics

1 answer:

liberstina [14]3 years ago

8 0

Answer:

4/5

Step-by-step explanation:

2x7-10/9-4

14-10/5

4/5

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Please solve equation<br> -3x2+243=0

Illusion [34]

Answer is 9.

-3x^2 + 243 = 0
Bring 243 to the right.
-3x^2 = -243
Divide by -3.
x^2 = 243/3 = 81
Tak the square root of both sides.
x = 81^1/2 = 9.

3 0

2 years ago

Read 2 more answers

Suppose we want a 90% confidence interval for the average amount spent on entertainment (movies, concerts, dates, etc.) by fresh

katovenus [111]

Answer:

609

Step-by-step explanation:

Standard deviation = $\sigma$ = $30

Margin of error = E = $2

Confidence level = 90%

Since the distribution is said to be normal, we will use z scores to solve this problem.

The z score for 90% confidence level = z = 1.645

Sample size= n = ?

The formula to calculate the margin of error is:

$E=z\frac{\sigma}{\sqrt{n}}\\\\\sqrt{n}=z\frac{\sigma}{E}\\\\n=(\frac{z\sigma}{E} )^{2}$

Using the values in above equation, we get:

$n=(\frac{1.645 \times 30}{2} )^{2}\\\\ n = 608.9$

This means, the minimum number of observations required is 609

6 0

3 years ago

PLS HELP ME

eimsori [14]

Answer:

5.5

Step-by-step explanation:

2,3,3,4,5,7,8,8,8,10,10,12

The mean of this data is 7.5. So what you are trying to figure out is the mean between 2 and 7. Since that is 3 and 4 it would be 3.5. So 3.5 is the Q1. Now you have to figure out the mean between 8 and 12. That is 9. This is Q3. You have to subtract 9-3.5 and that is the interquartile range. So 5.5 is the answer. I hope that made sense.

6 0

2 years ago

An instructor gives her class the choice to do 7 questions out of the 10 on an exam.

Maksim231197 [3]

Answer:

(a) 120 choices

(b) 110 choices

Step-by-step explanation:

The number of ways in which we can select k element from a group n elements is given by:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

$10C7=\frac{10!}{7!(10-7)!}=120$

Then each student have 120 possible choices.

On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:

1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:

$(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50$

2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:

$(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50$

3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:

$(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10$