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IrinaVladis [17]
2 years ago
6

What is the slope intercept formula​

Mathematics
2 answers:
Nataly_w [17]2 years ago
5 0

Answer:

y=mx+b

Step-by-step explanation:

Nutka1998 [239]2 years ago
4 0

Answer:

y=mx+b

Explanation:

The slope-intercept formula is y=mx+b. The b represents the y-intercept, which is the point where a line on a graph meets the y-axis. The coordinates for this would be (0,b). The m represents the slope, which is the rate of change. The x represents any x variable that you plug in. The y represents the y variable.

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What is value of x? <br> See picture
morpeh [17]

Answer:

x = 71

Step-by-step explanation:

Since 2 sides of the triangle are congruent then the triangle is isosceles.

Thus the 2 base angles are congruent

The sum of the 3 angles in the triangle = 180°, so

x = \frac{180-38}{2} = \frac{142}{2} = 71

6 0
3 years ago
Read 2 more answers
What is 80/400 as a decimal
elena55 [62]
80/400 as a decimal is 0.2
4 0
3 years ago
Tommy had a rectangular strip of paper that is 21 cm long. If he cuts it into smaller pieces, each 2 1/3 cm long, how many piece
abruzzese [7]

Answer:

The answer is 9.

Explanation: 21÷ 2 1/3= 9

7 0
2 years ago
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. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0
2 years ago
Please help right away
STALIN [3.7K]

Answer:

443 jelly boys

Step-by-step explanation:

the raius is 3.5, then square it is 12.25, multiplied by pie,  12.25*3.14159=38.4844775, then times 11.5 is 442.57149125‬, aproximately 443.

4 0
3 years ago
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