$350, 6.2%, 3 years plzz help

Brady rode his bike 70 miles in 4 hours. He rode at an average speed of 17 miles per hour for t hours and at an average rate of

IceJOKER [234]

Answer:

Brady rides at slower speed which is 17 miles/hours for <em><u>3 hours and 36 minutes.</u></em>

Step-by-step explanation:

Given:

Brady rode 70 miles in 4 hours.

For $t$ hours he rode at the average rate of 17 miles/hour

For rest of the time he rode at the average rate of 22 miles per hour.

To find the time Braady rode at the slower speed.

Solution:

Total time of riding = 4 hours

Time for which Brady rides at 17 miles/ hour = $t$ hours

Distance covered in $t$ hours = $Speed\times time=17\times t =17t\ miles$

So, time for which Brady rides at 22 miles/ hour = $(4-t)$ hours

Distance covered in $(4-t)$ hours = $Speed\times time=22\times(4-t) =(88-22t)\ miles$

Total distance can be given as:

⇒ $17t+88-22t$

Simplifying.

⇒ $-5t+88$

Total distance given =70 miles.

Thus, the equation to find $t$ can be given as:

$-5t+88=70$

Subtracting both sides by 88.

$-5t+88-88=70-88$

$-5t=-18$

Dividing both sides by -5.

$\frac{-5t}{-5}=\frac{-18}{-5}$

∴ $t=3.6$ hours

$3.6\ hours = 3\ hour+ (0.6\times 60)\ minutes = 3\ hour\ 36\ minutes$

Thus, Brady rides at slower speed which is 17 miles/hours for 3 hours and 36 minutes.

6 0

3 years ago

Find a common difference of the arithmetic sequence -2, -1 4/5, -1 3/5, -1 2/5,

elixir [45]

Answer:

The common difference is $\frac{1}{5}$ or 0.2

Step-by-step explanation:

Given:

Arithmetic Sequence -2, -1 4/5, -1 3/5, -1 2/5,

First term = a₁ = -2

Second term = a₂ = $\frac{-9}{5}$

Third term = a₃ = $\frac{-8}{5}$

Fourth term = a₄ = $\frac{-7}{5}$

To Find:

Common Difference = d = ?

Solution:

Formula for Common Difference d , $\textrm{common difference}= \textrm{Second Term} -\textrm{First Term}=\textrm{Third Term}-\textrm{Second Term}$

$\textrm{common difference} = a_{2}- a_{1}= a_{3}- a_{2}=a_{4}- a_{3}$

∴ $d = \frac{-9}{5}-(-2)\\d=\frac{-9}{5}+2\\d=\frac{1}{5}\\ \therefore d =0.2\\$

The common difference is $\frac{1}{5}$ or 0.2

6 0

3 years ago

Identify the points in Figure 2 that correspond to the points A and C in Figure 1. Label them p and r . What is the distance bet

Otrada [13]

Answer:

pls send the picture of the question

3 0

3 years ago

PLEASE HELP

lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function $j\left(x\right) \ = \ 4^{x-2}$ , hence the average rate of change (slope of the blue line) between the $x$ and $x+h$ is

$\text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}$

$\\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}$

7 0

1 year ago

Find the mean, median, and mode of the data with and without the outlier. 101, 110, 99, 100, 64, 112, 110, 111, 102

svetoff [14.1K]

Answer:

Step-by-step explanation:

First step

Put the numbers in numerical order.

64 99 100 102 110 110 111 112

Median

Median is the middle number. If the number of numbers is even then you take the middle 2 and average them.

This has 8 entries so you take the middle 2 and average them.

102 and 110 are the middle 2. They can be averaged.

(102 + 110)/2 = 212/2

212 / 2 = 106

The median is 106.

Mode

The mode is the entry with the most number that are the same.

110 is given twice. Nothing else is. So the mode is 110

Mean

The mean is the average. You add the eight numbers and divide by 8.

Sum = 64 + 99 + 100 + 102 + 110 + 110 + 111 + 112

Sum = 808

Now divide by 8

Mean = 808/8

Mean = 101

6 0

1 year ago