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egoroff_w [7]
3 years ago
14

Carlos biked StartFraction 139 over 8 EndFraction miles on Saturday and StartFraction 135 over 7 EndFraction miles on Sunday. On

which day did he ride further and by how much? Carlos rode further on Saturday by 1 and StartFraction 51 over 56 EndFraction miles. Carlos rode further on Saturday by 3 and StartFraction 5 over 56 EndFraction miles. Carlos rode further on Sunday by 3 and StartFraction 51 over 56 EndFraction miles. Carlos rode further on Sunday by 1 and StartFraction 51 over 56 EndFraction miles. Mark this and return
Mathematics
2 answers:
melomori [17]3 years ago
7 0

Answer:

Carlos rode further than on Saturday by 107/56 miles.

Step-by-step explanation:

i did the quiz lol

Marianna [84]3 years ago
6 0

Answer:

Its D

Step-by-step explanation:

I took the test and got it right.

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In the graph, if the number of manufactured units falls below 6.25, then the business will be operating in what state?
AURORKA [14]
It would be in a state of loss because the units manufactured neither reaches nor exceeds the break even point.
6 0
3 years ago
P=f/a, where f=70 and a=20 f is increased by 10 a is increased to by 10 <br><br>​
valina [46]

Answer:

P=3.5

P=2.667

Step-by-step explanation:

p=f/a

P=70/20

P=3.5

then f is increased by 10

f=70+10=80

and a is increased by 10

a=20+10

a=30

therefore; P=f/a

P=80/30

P=2.667

3 0
2 years ago
There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=
yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = \frac{12P4}{12^{4} }

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = \frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }

P(X=2) = 0.401

P(X=3) = \frac{4C3 * 12P2}{12^{4} }

P(x=3) = 0.025

P(X=4) = \frac{12}{12^{4} }

P(X=4) = 0.001

Mean, \mu = \sum xP(x)

\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879

Standard deviation, SD = \sqrt{\sum x^{2} P(x) - \mu^{2}}  \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2}  * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}

SD = 1.033

4 0
4 years ago
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts
Hunter-Best [27]

Answer:

The value of the test statistic is t = 2.1

Step-by-step explanation:

Our test statistic is:

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the expected value, \sigma is the standard deviation and n is the size of the sample.

The level of ozone normally found is 7.5 parts/million (ppm).

This means that \mu = 7.5

The mean of 24 samples is 7.8 ppm with a standard deviation of 0.7.

This means that X = 7.8, \sigma = 0.7, n = 24

Test Statistic:

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

t = \frac{7.8 - 7.5}{\frac{0.7}{\sqrt{24}}}

t = 2.1

The value of the test statistic is t = 2.1

7 0
3 years ago
What does the zero on a graph represent
nasty-shy [4]

Answer:

the origin

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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