Determine whether the given arc is a minor arc, major arc or semicircle.

The figure is made up of 2 rectangles and 2 right triangles.

Zarrin [17]

Answer:

area = 277 units²

Step-by-step explanation:

area = (25 x 5) + (8 x 13) + (8 x 6) = 277 units²

7 0

3 years ago

Mai and Priya were on scooters. Mai traveled 15 meters in 6 seconds. Priya traveled 22 meters in 10 seconds. Who was moving FAST

LenaWriter [7]

Answer:

mai is faster

Step-by-step explanation:

because even though she on goes 15 meters in 6 seconds she would have 4 times the advantage because of the extra 4 seconds she has

6 0

3 years ago

Hi<br>:):):):):):):):):):):)

Aleks [24]

Answer: HI!

Step-by-step explanation:

6 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

What is the solution to ther equation 2(x-4)squared=50

galina1969 [7]

Answer:

x=9 x=-1

Step-by-step explanation:

2(x-4)^2=50

Divide each side by 2

2/2(x-4)^2=50/2

(x-4)^2=25

Take the square root of each side

sqrt((x-4)^2)=sqrt(25)

x-4 = ±5

x-4 =5 x-4 = -5

Add 4 to each side

x-4+4=5+4 x-4+4 = -5+4

6 0

3 years ago