1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nutka1998 [239]
3 years ago
7

FREE POINTS AND MYBE A BRAINLIEST JUST ONE QUESTION ARE YOU TIRED OF ASKING SONE QUESTIONS NO ONE ANSWERS AND THEN YOU CAN'T GET

YOUR POINTS BACK BECAUSE I AM
Mathematics
2 answers:
natka813 [3]3 years ago
3 0

Answer:

Yes totally friend, I totally hear you on that!! ^-^ Good luck for your weekend!!!

Step-by-step explanation:

sdas [7]3 years ago
3 0

Answer:

yeah

sorry, i probably shouldn't be saying this, but you can also use quizlet or coursehero for homework help

:)

You might be interested in
Solve the equation 1/3m+6m-9/3m=3m-3/4m
vekshin1
M
=
0

O
r

m
=
23
15

Explanation:
1
3
m
+
6
m
−
9
3
m
=
3
m
−
3
4
m



⇒
1
+
6
m
−
9
3
m
=
3
m
−
3
4
m





⇒
−
8
+
6
m
3
m
=
3
m
−
3
4
m





⇒
(
−
8
+
6
m
)
×
4
m
=
(
3
m
−
3
)
×
3
m





⇒
−
32
m
+
24
m
2
=
9
m
2
−
9
m





⇒
−
32
m
+
24
m
2
−
9
m
2
+
9
m
=
0





⇒
15
m
2
−
23
m
=
0





⇒
m
(
15
m
−
23
)
=
0



m
=
0



Or


15
m
−
23
=
0
⇒
15
m
=
23
⇒
m
=
23
15
3 0
3 years ago
Order of operations 20/5*2
Xelga [282]
The order of operations is Parenthesis Exponents Multiplication Division Addition Subtraction For this problem, we have M and D. We first do division. Not always multiplication is first. So we divide 20 by 5, which gets you 4, and multiply by 2, which is 8. 8 IS YOUR FINAL ANSWER,
6 0
3 years ago
Read 2 more answers
The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are 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.

In this question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

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

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

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

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0
3 years ago
There are 5 blue
Minchanka [31]

Answer:

The probability of randomly picking a red candy is 1/20 and In percentage value 5%

Step-by-step explanation:

There are 5 blue candies, 6 red candies, and 9 yellow candies.

Now, Total

5 + 6 + 9 = 20 candies

Now, The probability of randomly picking a red candy is 1/20

Now, In percentage

1/20 × 100% = 5%

Thus, The probability of randomly picking a red candy is 1/20 and In percentage value 5%.

<u>-TheUnknownScientist</u>

5 0
3 years ago
Is 83 a prime or composite number? <br> Please help me out....
adoni [48]
To find out if a number is prime or composite, divide the number by 2. If it gives no remainder, it is a composite number. Otherwise, it's a prime number.

83 / 2 = 41 R 1
It has a remainder, so it's a prime number.
5 0
3 years ago
Read 2 more answers
Other questions:
  • How to solve -6x + 5y=-6 -3y
    6·1 answer
  • The measure of the supplement of an angle is seven times as large as the measure of its complement find the measure of the angle
    15·1 answer
  • What is the inequality shown?
    10·1 answer
  • Do u simplify 7 wholes and 8/10
    9·1 answer
  • I need to solve the equation please help me
    8·1 answer
  • Tony borrows $7,000 from a friend
    9·2 answers
  • 1/2 or (1/2)² which value is greater​
    8·1 answer
  • Please help me answer this question?
    13·2 answers
  • NEED HELP it’s a test please some one
    14·1 answer
  • A pole that is 2.6 m tall casts a shadow that is 1.26 m long. At the same time, a nearby tower casts a shadow that is 37.25 m lo
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!