Yuri has a rope 12m long. he cuts it into pieces that are each 2m long. what fraction of the rope is one piece

Which value for x makes the sentence true?

Nat2105 [25]

A. x = 2
Would make your statement true

3 0

3 years ago

Help me answer this question about congruent triangles. 50 points available.

Anuta_ua [19.1K]

Answer:

B

Step-by-step explanation:

No these triangles are not congruent.

Left triangle

Shortest side = 6 cm

Longest side = 13 cm

3rd side = unknown but < 13

Right triangle

Shortest side = 6 cm

Longest side = unknown but > 13

3rd side = 13 cm

Although the shortest side of both triangles is 6 cm, the longest side of the left triangle is 13 cm, whereas the longest side of the right triangle is unknown but will be more than 13 cm.

We do not know if any of the angles are congruent. If they were congruent, we would expect to see this marked by the same angle line(s) on each triangle.

3 0

2 years ago

The image of a dust mite from a scanning electron microscope is 1.5×10 to the 2nd millimeters wide. The image is 5×10 to 2nd tim

Vinil7 [7]

600 millimeters is the accuracy

7 0

2 years ago

Read 2 more answers

A circle with a radius of 5 cm has an area of 78.54 cm2. What constant, rounded to the nearest hundredth, do you get when the ar

Naily [24]

Answer:

3.14

Step-by-step explanation:

Area of a corcle is sexpressed as';

A = πr²

π = A/r²

π = 78.54/5²

π = 78.54/25

π = 3.1416

When the area is divided by radius the required constant is 3.14

3 0

3 years ago

A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while

ki77a [65]

Answer:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=12$ represent the population standard deviation

n represent the sample size

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

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

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=1-0.96 =0.04$ . And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got $z_{\alpha/2}=2.054$ , replacing into formula (b) we got:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

5 0

3 years ago