If the simple interest on $6,000 for 10 years is $5,400, then what is the interest rate?

A dinner plate has a circumference of 106.76 cm. What is the area of the dinner plate? Use = 3.14.

Sphinxa [80]

C=2pr, r=c/(2p)

a=pr^2, using r found above we get:

a=p(c^2/(4p^2))

a=(c^2)/(4p), since c=106.76 and we approximate pi as 3.14

a=(106.76^2)/(4*3.14)

a=11612.2176/12.56 cm^2

a≈924.54 cm^2 (to nearest one-hundredth of a square cm)

3 0

3 years ago

Please help me with these !!! ❤️❤️ I need it thank youuu!

lisov135 [29]

For number 4, we'll need a few facts to answer our question:

- Two supplementary angles add up to 180°, forming a straight angle (the angle formed by a straight line)
- The interior angles of a triangle add up to 180°

Given those, we notice that the one unlabeled angle in the figure shares a line with 156°. In fact, this angle is supplementary to 156°, which means that the two add up to 180°. To find the measure of this mystery angle, we can subtract 156 from 180 to obtain 180 - 156 = 24°.

Now, let's look at the triangle. We already know the measure of one of the angles is 24°, and the other two are x°. What else do we know about the angles of a triangle? From the two facts listed at the beginning, we know their interior angles add up to 180°, so let's use that fact to solve for x.

We have:

$x+x+24=180$ , or $2x+24=180$

Solving for x:

$(2x+24)-24=180-24\\2x=156\\(2x)/2=(156)/2\\x=78$

So, x = 78°.

For question 5, the definition of a pair of parallel lines is a pair of lines which never intersect, so "always" would be the appropriate answer.

7 0

3 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0

3 years ago

Help please! please make the answers original.

baherus [9]

Answer:

I will do it but what is the GCF of part B

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

In an area of the Midwest, records were kept on the relationship between the rainfall (in inches) and the yield of wheat (bushel

dusya [7]

Answer:

The correct answer is 36.2

Step-by-step explanation:

Hello!

Given the data:

Rainfall(inch) x: 10.5, 8.8, 13.4, 12.5, 18.8, 10.3, 7.0, 15.6, 16.0

Yield (bushels per acre) y: 50.5, 46.2, 58.8, 59.0, 82.4, 49.2, 31.9, 76.0, 78.8

The response variable is

Y: Yield of wheat (bushels per acre)

Explanatory variable:

X: Rainfall in a certain period (inch)

I've calculated the equation of regression using a statistics software, the estimated equation is:

^Y= 4.27 + 4.38X

To calculate the value that will take the response variable for a given value of X:

^Y= 4.27 + 4.38*7.3= 36.24 bushels per acre.

I hope it helps!

5 0

3 years ago