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Ronch [10]
3 years ago
15

I have 2 Questions.

Mathematics
2 answers:
Nesterboy [21]3 years ago
6 0
What type of class are you in
aleksandr82 [10.1K]3 years ago
5 0

Answer:

looks hard omgong

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If a= -9 and b= -4, what is the value of a + b?
Anna007 [38]

Answer:

-13

Step-by-step explanation:

a=-9 and b= -4

-9 + -4=

-13

HOPE IT HELPS!!!!

PLEASE MARK BRAINLIEST!!

4 0
4 years ago
Read 2 more answers
How to find the answer
KATRIN_1 [288]
Use this equation:
A = P(1+r)^t

Where A is the final amount
P is the initial amount
r is the annual rate
t is the time in years

P = 1500
r = 0.07
t = 3

A = 1500 (1.07^3)
≈ $1837.56

Have an awesome day! :)
5 0
3 years ago
Solve Az + By = C for y
Viefleur [7K]

Answer:

y= (C - Az)/B

Step-by-step explanation:

Az +By = C

Az - Az + By = C - Az

By = C - Az

Divide both side by the coefficient of y which is B

therefore we have:

y = (C - Az)/B

7 0
2 years ago
Read 2 more answers
Solve : |1/5x - 7 | =4
Mnenie [13.5K]

Answer:

x = 55 or x = 15

Step-by-step explanation:

Solve Absolute Value.

We know that either 1/5x - 7 = 4 or 1/5x - 7 = -4.

1/5x - 7 = 4

1/5x - 7 + 7 = 4 + 7

1/5x = 11

1/5x (5) = 11 (5)

x = 55 (one possibility)

1/5x - 7 = -4

1/5x - 7 + 7 = -4 + 7

1/5x = 3

1/5x (5) = 3 (5)

x = 15 (second possibility)

6 0
4 years ago
Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f
schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval:  (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{130}{20} = 6.5

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

S.D = \sqrt{\frac{233.5}{19}} = 3.50

b) Standard Error

= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826

c) Point estimate for the mean time required for all college is given by the sample mean.

\bar{x} = 6.5

d) 90% Confidence interval:  

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729  

6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0
4 years ago
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