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Paul [167]
1 year ago
9

Evaluate. 75-2(9-5)²+2³

Mathematics
1 answer:
Andreyy891 year ago
5 0
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

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Find the total surface area of a cuboid 7.5cm 2.3cm 5cm
zavuch27 [327]

Answer:

17.25

Step-by-step explanation:

im not sure with the answer without a photo.formula to find area is LENGTH×WIDTH so there u can find it

6 0
3 years ago
Read 2 more answers
I NEED HELP THIS IS LATE
slamgirl [31]

Answer:

M=-10

Step-by-step explanation:

To find the sope we use y-y1/x-x1. For this problem, I chose the first two coordinates.

(-33.5)-(-55.6)/1.20-3.45=22.6/-2.25=-10(rounded)

hope this helps!:)

6 0
3 years ago
Which ordered pairs lie on the graph of the exponential function f(x)=4(5)2x?
Katen [24]
(-1, 4/25) and (2,2500)

4 0
3 years ago
A group of friends are playing a trivia game. players spin a spinner 2 times to find the 2 categories of trivia questions they w
Delicious77 [7]

The probability that a player will be asked a math question and then a music question is 0.03125

<h3>How to determine the probability?</h3>

The size of the sections are given as:

Music = 0.5

Sport = 0.5

Others = 1

So, the total size is:

Total = 0.5 + 0.5 + 1 + 1 + 1

Evaluate

Total = 4

The probability of asking a math question is:

P(Math) = 1/4 = 0.25

The probability of asking a music question is:

P(Music) = 0.5/4 = 0.125

The required probability is:

P(Math and Music) = P(Math) * P(Music)

This gives

P(Math and Music) = 0.25 * 0.125

Evaluate

P(Math and Music) = 0.03125

Hence, the probability that a player will be asked a math question and then a music question is 0.03125

Read more about probability at:

brainly.com/question/25870256

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5 0
2 years ago
A random sample of 85 group leaders, supervisors, and similar personnel at General Motors revealed that, on average, they spent
andreyandreev [35.5K]

Answer:

Step-by-step explanation:

From the information given,

Number of personnel sampled, n = 85

Mean or average = 6.5

Standard deviation of the sample = 1.7

We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.

For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.

z×standard deviation/√n

= 1.96 × 6.5/√85

= 1.38

The confidence interval for the mean number of years spent before promotion is

The lower end of the interval is 6.5 - 1.38 = 5.12 years

The upper end is 6.5 + 1.38 = 7.88 years

Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years

4 0
3 years ago
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