Smb help w this pls asap

A parallelogram is drawn on a

mezya [45]

Answer:

(-3 , -3)

Step-by-step explanation:

A: (-1,2), B: (1,2), and C: (-1,-3) .... D: (x,y)

slope of AB = (2-2)/(1 - -1) = 0 .... parallel to x axis

x coordinate of A is 2 units left to B (-1 vs 1)

y coordinate of C is 2 units left to B so D must be 2 units left to C: -1 -2 = <u>-3</u>

DC // AB , y coordinate of D = C =<u> -3</u>

D: (-3 , -3)

6 0

3 years ago

State that the triangles in each pair of similar. If so, complete the similarity statement, and find the scale factor for each o

ser-zykov [4K]

9514 1404 393

Answer:

ΔLMN ~ ΔLFG; scale factor 7/3

Step-by-step explanation:

The ratios of side lengths, shortest to longest are ...

FG : GL : LF = 24 : 36 : 42 = 4 : 6 : 7 (reduced)

MN : NL : LM = 56 : 84 : 98 = 4 : 6 : 7 (reduced)

The side ratios are the same, so ΔLMN ~ ΔLFG. The scale factor is ...

MN/FG = 56/24 = 7/3

7 0

3 years ago

Read 2 more answers

How do you solve a system of equation?

Eva8 [605]

Just subtract and then add then you should get your answer.

3 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 526}{47.86}$

$Z = -2.42$

$Z = -2.42$ has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0

3 years ago

Simplify the expression shown below: (a^3b^12c^2)^2(a^5c^2)^3(b^5c^4)^0

NemiM [27]

= a^21b^24c^10
use the product rules for each set of () then multiply them together. when multiplying variables with exponents you just add them together like for a^3 * a^5 = a^8. (a^3 b^12 c^2)^2 = (a^6 b^24 c^4)

4 0

3 years ago