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luda_lava [24]
2 years ago
12

- Vicki bought 5 vegetables for every 7 fruitsshe bought.

Mathematics
1 answer:
andrezito [222]2 years ago
7 0

Answer:

5:7

Step-by-step explanation:

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Answer:

the correct graph is b for this problem

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3 years ago
4. Write an equation for the table in slope intercept form​
Hunter-Best [27]

Answer:

y=1.6x-5

Step-by-step explanation:

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2 years ago
Over the last 6 days, you earned the following amounts raking leaves: $44, $34, $54, $55, $61, and $37. On the average, how much
astra-53 [7]

Answer: $47.50

Step-by-step explanation: Averages are calculated by adding all values up and dividing by # of values. Here the sum is 285, and 285/6=47.5

6 0
3 years ago
a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was
Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em>\bar X<em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

          Z = \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean hours spent studying = 25 hours

            \sigma = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < \bar X < 30 hours)

    P(29 hours < \bar X < 30 hours) = P(\bar X < 30 hours) - P(\bar X \leq 29 hours)

      

    P(\bar X < 30 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } < \frac{ 30-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z < 2) = 0.97725

    P(\bar X \leq 29 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 29-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z \leq 1.60) = 0.94520

                                                                    

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < \bar X < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0
2 years ago
The cost to write an ad in a news paper is $10 plus $0.25 per word. Write an inequality that can be used to find the maximum amo
MA_775_DIABLO [31]

Answer:

f(x)=10+0.25x

she can put maximum 20 words

Step-by-step explanation:

f(x)<=15

10+0.25x<=15

0.25x<=5

x<=20

8 0
3 years ago
Read 2 more answers
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