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vfiekz [6]
3 years ago
7

Emma is training for a 10-kilometer race. She wants to beat her last 10-kilometer time, which was 1 hour 10 minutes. Emma has al

ready run for 55 minutes. Which inequality can be used to find how much longer she can run and still beat her previous time? (1 hr = 60 min)
70 greater-than 7 minus 55
70 less-than t minus 55
70 less-than-or-equal-to t + 55
70 greater-than t + 55
x + 6 less-than negative 8

x + 4 greater-than-or-equal-to negative 6

x minus 3 greater-than negative 10

x + 5 less-than-or-equal-to negative 4

A
B
C
D
Mathematics
2 answers:
Mandarinka [93]3 years ago
7 0

Answer:

70 > 55 + t

Step-by-step explanation:

Given

Previous\ Time = 1\ hr\ 10\ mins

Current\ Time = 55\ mins

Required

Represent the additional time needed as an inequality

Convert the previous time to minutes

Previous\ Time = 1\ hr\ 10\ mins

Previous\ Time = 1 * 60\ mins + 10\ mins

Previous\ Time = 60\ mins + 10\ mins

Previous\ Time = 70\ mins

Let the additional time be t

<em>This means that she needs to run at least t minutes to beat her previous record</em>

<em />

To beat her previous record, the sum of current time and x must be less than her previous time. i.e.

55 + t < 70

This can be rewritten as:

70 > 55 + t

Mazyrski [523]3 years ago
4 0

Answer:the answer is D

Step-by-step explanation:

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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%

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4 years ago
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Answer:

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Step-by-step explanation:

Step  1  :

           h23

Simplify   ———

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Equation at the end of step  1  :

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 ((f9) • ———) • h17

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Step  2  :

Multiplying exponential expressions :

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Final result :

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Answer

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