Can somebody help me on question 15?!

A system of equations has infinitely many solutions. If zy - 4x = 6 is one of the equations, which could be the other equation?

NemiM [27]

Answer:

I am not fully sure but C

4 0

2 years ago

Help me please!!

Setler [38]

Answer:

3.A

2.59 %

3.B

.16 x 2,000 = 320

6 0

3 years ago

In Exercises 26 and 27 write an equation of the line with the given slope and y-intercept.

Neko [114]

26. y= -7/3+ 0 or just y= -7/3

27. y= -10

7 0

2 years ago

The area of a square rug is 169ft^2. How long is each side of the rug?

jek_recluse [69]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{13 \: ft}}}}}$

Step-by-step explanation:

$\star{ \sf{ \: Area \: of \: a \: square \: rug \: ( \: A \: ) \: = \: 169 \: {ft}^{2} }}$

$\sf{ \underline{Finding \: the \: length \: of \: a \: square \: ( \: l \: )}}$

$\boxed{ \sf{Area \: of \: a \: square \: = \: {l}^{2} }}$

$\hookrightarrow{ \sf{169 = {l}^{2} }}$

$\hookrightarrow{ \sf{ {l}^{2} = 169}}$

$\hookrightarrow{ \sf{ \sqrt{ {(l)}^{2} } = \sqrt{169} }}$

$\hookrightarrow{ \text{length \: = \: 13 \: ft}}$

$\text{Hope \: I \: helped!}$

$\text{Best \: wishes !!}$

~ $\text{TheAnimeGirl}$

6 0

2 years ago

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.

devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0

3 years ago