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Crazy boy [7]
3 years ago
6

PLEASE HELP!!!!! I WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

Mathematics
1 answer:
Luda [366]3 years ago
4 0

Answer:

Step-by-step explanation:

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Can some one help me w my math
KiRa [710]

Answer: It would be B

Step-by-step explanation:

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7 0
3 years ago
Help me plz these are my last points and my las try the answer i did said its wrong
irga5000 [103]

Answer:

b=8h, not 8/h

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4 0
3 years ago
Read 2 more answers
4. Which graph represents the solution to<br> -2&lt;3x+1 ?<br>​
dalvyx [7]

Hey there!

-2<3x+1

= 3x + 1 < -2

SUBTRACT 1 to BOTH SIDES

3x + 1 - 1 < -2 + 1

CANCEL out: 1 - 1 because that gives you 0

KEEP: -2 + 1 because that helps you solve for x

-2 + 1 = -3

New EQUATION: 3x > -3

DIVIDE 3 to BOTH SIDES

3x/3 > -3/3

CANCEL out: 3/3 because that gives you 1

KEE3 -3/3 because that helps us compare for x

-3/3 = -1

Answer to the equation: x > -1

OVERALL Answer: x > -1; It is an OPENED circle shaded to the right starting at -1 (Option A.)

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

8 0
3 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
Which of the following is equivalent to 5 cube root 13^3
shtirl [24]

Answer:

d. 13 raise power 3/5

Step-by-step explanation:

3 0
3 years ago
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