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3 years ago

6

Tanya is training a turtle for a turtle race. For every of an hour that the turtle is crawling, he can travel of a mile. At what

unit rate is the turtle crawling?

2 answers:

just olya [345]3 years ago

8 0

FOR EVERY MILE, IT TAKES AN HOUR...
SO THE UNIT RATE IS 1= MILES PER HOUR

lilavasa [31]3 years ago

7 0

The turtle is crawling at a rate of 1 mile per hour or 1 mph.

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Masja [62]

Answer:

(A) The rate of change in the price of a bushel of corn in the current year is $7.

(B) The price of a bushel of corn in the current year is $2 more than the price of a bushel of corn in the previous year.

Step-by-step explanation:

The graph for the prices of different numbers of bushels of corn at a store in the current year is shown below.

Part A:

The rate of change in the price of a bushel of corn in the current year based upon the number of bushels is known as the slope of the line.

The formula to compute the slope is:

$\text{Slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Consider the ordered pairs: (4, 28) and (10, 70)

Compute the slope of the line as follows:

$\text{Slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{70-28}{10-4}\\\\=\frac{42}{6}\\\\=7$

Thus, the rate of change in the price of a bushel of corn in the current year is $7.

Part B:

The data for the price of bushels in the previous year is as follows:

Number of Bushels Price

2 10

4 20

6 30

8 40

Compute the rate of change in the price of a bushel of corn in the previous year based upon the number of bushels as follows:

Consider the ordered pairs: (2, 10) and (6, 30)

$\text{Slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{30-10}{6-2}\\\\=\frac{20}{4}\\\\=5$

The rate of change in the price of a bushel of corn in the previous year is $5.

Thus, the price of a bushel of corn in the current year is $2 more than the price of a bushel of corn in the previous year.

5 0

3 years ago

Need help pleaseI was bad at math in school so lwant to learn

aleksklad [387]

The probability of an event is expressed as

$Pr(\text{event) =}\frac{Total\text{ number of favourable/desired outcome}}{Tota\text{l number of possible outcome}}$

Given:

$\begin{gathered} \text{Red}\Rightarrow2 \\ \text{Green}\Rightarrow3 \\ \text{Blue}\Rightarrow2 \\ \Rightarrow Total\text{ number of balls = 2+3+2=7 balls} \end{gathered}$

The probability of drwing two blue balls one after the other is expressed as

$Pr(\text{blue)}\times Pr(blue)$

For the first draw:

$\begin{gathered} Pr(\text{blue) = }\frac{number\text{ of blue balls}}{total\text{ number of balls}} \\ =\frac{2}{7} \end{gathered}$

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).

Thus,

$\begin{gathered} Pr(\text{blue)}=\frac{number\text{ of blue balls left}}{total\text{ number of balls left}} \\ =\frac{1}{6} \end{gathered}$

The probability of drawing two blue balls one after the other is evaluted as

$\begin{gathered} \frac{1}{6}\times\frac{2}{7} \\ =\frac{1}{21} \end{gathered}$

The probablity that none of the balls drawn is blue is evaluted as

$\begin{gathered} 1-\frac{1}{21} \\ =\frac{20}{21} \end{gathered}$

Hence, the probablity that none of the balls drawn is blue is evaluted as

$\frac{20}{21}$

8 0

1 year ago

Last week, Evelyn ran 44 laps at the community center. Peter ran 75% of the number of laps that Evelyn ran. How many laps did Pe

emmainna [20.7K]

Answer:

Peter ran 33 laps.

Step-by-step explanation:

Multiply Evelyn's 44 laps by 75% or 0.75 to get your answer.

44(0.75)=33

3 0

3 years ago

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

8 0

3 years ago

A hiking trail rises 9 feet for every horizontal change of 150 feet. What is the slope of the hiking trail?

Kipish [7]

The slope of the trail would be 18. hope it helps

3 0

3 years ago

Read 2 more answers