I need help understanding how to do this please ASAP!!!!!!

Evaluate. 75-2(9-5)²+2³

Andreyy89

$75-2(9-5)^2+2^3$

First, solve the parentheses:

$75-2(4)^2+2^3$

Next, the exponential terms

$75-2(16)+8$

Solve the multiplication:

$75-32+8$

Add and subtract:

$51$

5 0

1 year ago

What is the perimeter of the square?

KATRIN_1 [288]

The answer is A. 24 yards because to find perimeter, you add up the lengths of all four sides

6 0

3 years ago

Read 2 more answers

Shade in some or all of the grids to show that 2 ones is the same as 20 tenths

scZoUnD [109]

Given three grids with each grid representing one whole.
Each grid represents 1 ones and each vertical bar of a grid represent 1 tenth.
There are 10 vertical bars in each grid, hence each grid represent 1 ones = 10 tenths and 2 grids represent 2 ones = 20 tenths.

Therefore, two complete grids are to be shaded to show that 2 ones is the same as 20 tenths

4 0

3 years ago

Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p

scZoUnD [109]

Answer:

The probability that they will teach different courses is $\frac{2}{3}$ .

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here x represents the course taught by the first part-time instructor and y represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events E is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Compute the probability that they will teach different courses as follows:

$P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}$

Thus, the probability that they will teach different courses is $\frac{2}{3}$ .

3 0

3 years ago

According to a study conducted by the Toronto-based social media analytics firm Sysomos, of all tweets get no reaction. That is,

Svetllana [295]

Answer:

a) E(X) = 71

b) V(X) = 20.59

Sigma = 4.538

Step-by-step explanation:

The question is incomplete:

According to a 2010 study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January 5, 2015).

Suppose we randomly select 100 tweets.

a) What is the expected number of these tweets with no reaction?

b) What are the variance and standard deviation for the number of these tweets with no reaction?

This can be modeled with the binomial distribution, with sample size n=100 and p=0.71, as the probability of no reaction for each individual tweet.

The expected number of these tweets with no reaction can be calcualted as the mean of the binomial random variable with these parameters:

$E(X)=n\cdot p=100\cdot 0.71=71$

The variance for the number of these tweets with no reaction can be calculated as the variance of the binomial distribution:

$V(X)=np(1-p)=100\cdot0.71\cdot0.29=20.59$

Then, the standard deviation becomes:

$\sigma=\sqrt{V(X}=\sqrt{20.59}=4.538$

5 0

3 years ago