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igomit [66]
3 years ago
15

What number best completes both equations? 2/3÷3/10=_ _×3/10=2/3

Mathematics
1 answer:
dezoksy [38]3 years ago
7 0
The answer would be 20/9
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Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds a
kirill115 [55]

Answer:

12 feet per second.

Step-by-step explanation:

Please consider the complete question.

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be representedh(t)=-16t^2+20t.

What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?  

We will use average rate of change formula to solve our given problem.

\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}

\text{Average rate of change}=\frac{h(\frac{1}{2})-h(0)}{\frac{1}{2}-0}

\text{Average rate of change}=\frac{-16\cdot(\frac{1}{2})^2+20\cdot \frac{1}{2}-(-16\cdot(0)^2+20\cdot (0))}{\frac{1}{2}-0}

\text{Average rate of change}=\frac{-16\cdot\frac{1}{4}+10-(0)}{\frac{1}{2}}

\text{Average rate of change}=\frac{-4+10}{\frac{1}{2}}

\text{Average rate of change}=\frac{6}{\frac{1}{2}}

\text{Average rate of change}=\frac{2\cdot 6}{1}

\text{Average rate of change}=12

Therefore, Spot's average rate of ascent is 12 feet per second.

3 0
3 years ago
Find the area of the rectangle shown.<br> 38<br> 15<br> 15<br> 38
ki77a [65]

Answer:

324900

Step-by-step explanation:

7 0
3 years ago
F(3) =-2x^2+9x+5 calculate the following
ElenaW [278]

Answer:

50

Step-by-step explanation:

2(3)^2+9(3)+5

2(9)+27+5

18+27+5

=50

8 0
3 years ago
A student considers how mnay people enter a particular bank during the lunch hour and how many exit. She decided to run a simula
trapecia [35]

Answer:

The number of persons entering the bank based on given 20 simulations are 12

The number of persons exiting the bank based on given 20 simulations are 8

We have to find the probability that she will see a person entering the bank.

The probability that she will see a person entering the bank is:

\frac{12}{20}

0.6

Therefore, the option 0.600 is correct.


3 0
3 years ago
Read 2 more answers
The test scores on a 100-point test were recorded for 20 students:71 93 91 86 7573 86 82 76 5784 89 67 62 7277 68 65 75 84a. Can
Dafna11 [192]

Answer: a. Yes

              b. mean = 76.65

                  standard deviation = 10.04

              c. 76.65 ± 4.4

Step-by-step explanation:

a. <u>Stem</u> <u>and</u> <u>leaf</u> <u>Plot</u> shows the frequencies with which classes of value occur. To create this plot, we divide the set of numbers into 2 columns: <u>stem</u>, the left column, which contains the tens digits; <u>leaf</u>, the right column, which contains the unit digits.

<u>Normal</u> <u>distribution</u> is a type of distribution: it's a bell-shaped, symmetrical, unimodal distribution.

A stem and leaf plot displays the main features of the distribution. If turned on its side, we can see the shape of the data.

The figure below shows the stem and leaf plot of the 100-point test score. As we can see, when turned, the plot resembles bell-shaped distribution. So, this test scores were selected from a normal population.

b. <u>Mean</u> is the average number of a data set. It is calculated as the sum of all the data divided by the quantity the sample has:

mean = \frac{\Sigma x}{n}

For the 100-point test score:

mean = \frac{71+93+91+...+65+75+84}{20}

mean = 76.65

<u>Standard</u> <u>Deviation</u> determines how much the data is dispersed from the mean. It is calculated as:

s=\sqrt{\frac{\Sigma (x-mean)^{2}}{n-1} }

For the 100-point test score:

s=\sqrt{\frac{[(71-76.65)+(93-76.65)+...+(84-76.65)]^{2}}{20-1} }

s = 10.04

The mean and standard deviation of the scores are 76.65 and 10.04, respectively.

c. <u>Confidence</u> <u>Interval</u> is a range of values we are confident the real mean lies.

The calculations for the confidence interval is

mean ± z\frac{s}{\sqrt{n} }

where

z is the z-score for the 95% confidence interval, which is equal 1.96

Calculating interval

76.65 ± 1.96.\frac{10.04}{\sqrt{20} }

76.65 ± 4.4

The 95% confidence interval for the average test score in the population of students is between 72.25 and 81.05.

7 0
2 years ago
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