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3 years ago

9

What is the area of this parallelogram?

1 answer:

Dvinal [7]3 years ago

4 0

Answer:52cm

Step-by-step explanation:

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Oliver runs 23 fewer laps than nate. Nate runs 61 laps. How many laps does oliver run?

Crazy boy [7]

Answer:

38

Step-by-step explanation:

Since Oliver ran 23 fewer laps, you would use subtraction.

So you do 61-23=38

Hope this helps!!

Let me know if you are confused.

5 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through (–1, 5) and has a slope of 3 in point-slope form.

Mama L [17]

The point-slope form:

$y-y_1=m(x-x_1)$

We have:

$(-1,\ 5)\to x_1=-1,\ y_1=5\\m=3$

Substitute:

$y-5=3(x-(-1))\\\\y-5=3(x+1)$

Answer: C. y - 5 = 3(x + 1)

5 0

3 years ago

The diameter of a circle is 16 centimeters. How would you calculate the area?

Y_Kistochka [10]

Answer: 201.06

Step-by-step explanation:

To find the area of a circle with the radius, square the radius, or multiply it by itself. Then, multiply the squared radius by pi, or 3.14, to get the area. To find the area with the diameter, simply divide the diameter by 2, plug it into the radius formula, and solve as before.

Hope this helps!

6 0

3 years ago

Solve for x (13x + 2) (5x - 7) (3x - 4)<br><br><br> Thank you

Law Incorporation [45]

$\huge \bf༆ Answer ༄$

As we know, sum of interior angles of a triangle is 180°

that is ~

$\sf13x + 2 + 5x - 7 + 3x - 4 = 180$

$\sf21x - 9 = 180$

$\sf21x = 180 + 9$

$\sf x = \dfrac{189}{21}$

$\sf x = 9 \degree$

Hence, value of x is 9°

5 0

2 years ago

Read 2 more answers

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Vikentia [17]

Answer:

a

$\= x = 18.5$ , $\sigma = 5.15$

b

$15.505 < \mu < 21.495$

c

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

8 0

3 years ago