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marshall27 [118]
2 years ago
8

The Oregon Trail is 2,197 miles long. How long would it take a covered wagon traveling 20 miles a day to complete the trip? Writ

e the answer as a mixed number.
Mathematics
2 answers:
Valentin [98]2 years ago
7 0
It would take 109 17/20 of a day
Svetlanka [38]2 years ago
5 0

Answer:

109 17/20

Step-by-step explanation:

The distance is given as 2197 miles long at a speed of 20 miles per day. Speed is defined as distance over time:

V=d/t

We can calculate t because we have V and d:

20=2197/t

t=2197/20[tex][tex]t=109.85

109.85 in a mixed number is 109 85/100. We can divide the fraction by 5 we get 109 17/10.

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Which polynomials are in standard form?
Leni [432]

Answer:

B:x^4-8x^2-16

Step-by-step explanation:

B: x^4-8x^2-16 is in standard form.

4 0
2 years ago
Plz help i know you are smart
stellarik [79]

Answer:

The numbers on the axis need to follow a repeating pattern

I think it's the last one coz in the graph, 50 jumps to 58 which breaks the repeating rule of 5

7 0
3 years ago
The quotient of a number and 2 is larger than 3
Lostsunrise [7]
There are many for this ,but one sample is: 2 divided by 2is 4 X 3 which is 12.
8 0
2 years ago
HELP ASAP!!<br>The equation (blank) has no solution​
KengaRu [80]

Answer:

Just to recap, an equation has no solution when it results in an incorrect "equation".

For  example:

Equation: x+3 = x+4

Subtract x: 3 = 4???

But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.

Now onto our problem:

13y+2-2y = 10y+3-y

11y+2 = 9y+3

2y=1

y=1/2

9(3y+7)-2 = 3(-9y+9)

27y+61 = -27y+27

54y = -34

y = -34/54

32.1y+3.1+2.4y-8.2=34.5y-5.1

34.5-5.1=34.5y-5.1

5.1=5.1

infinite solutions

5(2.2y+3.4) = 5(y-2)+6y

11y+17 = 11y-10

17 = -10??

That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.

Let me know if this helps

6 0
2 years ago
A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed
tresset_1 [31]

Answer:

a) z= \frac{34-34}{2.5}= 0

z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

c) z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

X \sim N(\mu = 34, \sigma=2.5)

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

P(34 < X

We can find the number of deviation from the mean with the z score formula:

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

And replacing we got

z= \frac{34-34}{2.5}= 0

z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

P(29 < X

And replacing we got:

z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

7 0
2 years ago
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