I went to the mall with $75 and bought a pair of shoes and a few candy bars for $1.25 each. How much did I spend?

Cara is building a new dog house the roof is 10 4/5 feet long express this length as a decimal show work

scZoUnD [109]

Answer:

10.8 feet

Step-by-step explanation:

Multiply 4/5 by 2 on the top and bottom. Then you'll get 8/10 which is 0.8 as a decimal

7 0

2 years ago

Read 2 more answers

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.

Let $\bar X$ = sample average time spent studying

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

So, in the z table the P(Z $\leq$ 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.

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

Can anyone help me? I'm not getting anywhere!!

hram777 [196]

To analyze [insert whatever box and whisker plot does] and compare.

7 0

2 years ago

Tricia recorded the number of pets owned by each of her classmates. These data points represent the results of her survey. 0, 3,

mars1129 [50]

Answer:

first put them in order: 0,0,0,0,0,0,1,1,1,1,1,1,1,2,2,2,2,3,3,3,3,4,4

Step-by-step explanation:

then make a number like with the numbers 0 1 2 3 4 on it and put the amount of dots there are for that number

for example under 0 would be 6 0's because there are 6 0's

4 0

3 years ago

Read 2 more answers

Y=-0.05x^2+4.5x-6.5 pls help to solve

givi [52]

Following with the extra hint that x = 82.03 m

There are 2 ways to approach the question, but although I prefer to group up x first before plugging value, having that -0.05x^2 to group up is too annoying:

Plug x = 82.03

=> Y = -0.05*(82.03^2) + 4.5*82.03 - 6.5
= -336.446045 + 369.135 - 6.5
= 26.188955 m

Please take the time and understand the answer instead of just looking at the final result.

Also I know the number is very long, but that's the question's fault. ¯\_(ツ)_/¯

6 0

2 years ago