Where is the line of Reflection?

Solve the formula d = rt for t. A. t = rd B. t = d – r C. t = D. t =

tekilochka [14]

$rt=d\qquad\text{divide both sides by}\ r\neq0\\\\\dfrac{rt}{r}=\dfrac{d}{r}\\\\\boxed{t=\dfrac{d}{r}}$

3 0

3 years ago

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

77julia77 [94]

Answer:

x = 2 1/4 or 2.25

Step-by-step explanation:

to start we can make an equation

9lbs = 4x

divide by 4

9/4 = x

x = 2.25 or 2 1/4

4 0

3 years ago

What’s the digit you need to input in the space with the question mark to complete the code for the safe? Here’s a hint to save

elena55 [62]

Answer:

4

Step-by-step explanation:

2×2=4

6÷2=3

1×2=2

8÷2=4

3×2=6

?÷2=2

2×2=4

6 0

3 years ago

Read 2 more answers

a theater is designed with 15 seats in the first row, 19 in the second, 23 in the third, and so on. if this seating pattern cont

marusya05 [52]

Answer:

<u>131 seats</u> are in the 30th row.

Step-by-step explanation:

The theater is designed with the first row there are 15 seats, in second row 19 seats and in the third row there are 23 seats.

Now, to find the number of seats in the 30th row.

So, we get the common difference( $d$ ) from the arithmetic sequence first:

$19-15=4.$

Thus, $d=4.$

So, the first tem $a(1)$ = 15.

The number of last row ( $n$ ) = 30.

Now, to get the number of seat in the 30th row we put formula:

$a(n) = a(1) + d(n-1)$

$a(30)=15+4(30-1)$

$a(30)=15+4\times 29$

$a(30)=15+116$

$a(30)=131$

Therefore, 131 seats are in the 30th row.

4 0

4 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago

Where is the line of Reflection?​

Where is the line of Reflection?