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3 years ago

Isabella earns $48.75 for 5 hours of babysitting. At this rate, how much more would she earn for 11 hours of babysitting?

2 answers:

5 0

Isabella earned $48.75 for 5 hours of babysitting. If she babysits for 11 hours just take $48.75 and divide it by 5. Isabella earns $9.75 for every hour she babysits. Now multiply $9.75 by 11 and you should get $107.25. i hope this helped!

MAVERICK [17]3 years ago

4 0

She would earn $107.25

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-11r ≥ 220<br> how to solve? and what are the steps to solve<br> first gets marked brainliest

Bezzdna [24]

Step-by-step explanation:

here

-11r ≥ 220

-11r/-11 ≥ 220/-11 (dividing both side with -11 to eliminate -11 )

r ≥-20

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2 years ago

What is the estimate of 726

Nata [24]

To the nearest ones, it's 726.

To the nearest tens, it's 730.

To the nearest hundreds, it's 700.

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4 years ago

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Assume the heights in a female population are normally distributed with mean 65.7 inches and standard deviation 3.2 inches. Then

siniylev [52]

Answer:

0.620

Step-by-step explanation:

We know that 1 feet = 12 inches, so, 5 feet is equivalent to 60 inches. Then, we are looking for the probability that a typical female from this population is between 60 inches and 67 inches. We know that

$\mu$ = 65.7 inches and

$\sigma$ = 3.2 inches

and the normal density function for this mean and standard deviation is

$\frac{1}{\sqrt{2\pi } 3.2}exp[-\frac{(x-65.7)^{2}}{2(3.2)^{2}} ]$

The probability we are looking for is given by

$\int\limits^{67}_{60} {\frac{1}{\sqrt{2\pi } 3.2}exp[-\frac{(x-65.7)^{2}}{2(3.2)^{2}} ] } \, dx =0.620$

You can use a computer to calculate this integral. You can use the following instruction in the R statistical programming language

pnorm(67, 65.7, 3.2)-pnorm(60, 65.7, 3.2)

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3 years ago

Point S is between points R and T on RT. Use the information to write an equation in terms of x. Then solve the equation and fin

sergeinik [125]

Answer:

RS=6

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Step-by-step explanation:

answer is above

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4 years ago

Carly stated, "All pairs of rectangles are dilations. " Which pair of rectangles would prove that Carly’s statement is incorrect

xenn [34]

The pair of rectangle that will prove Carly's statement incorrect is: Option 3: A rectangle with length 4 and width 3. A rectangle with length 3 and width 2.

<h3>How to find if a pair of figure is not dilated version of each other?</h3>

Dilation of a figure will leave its sides get scaled (multiplied) by same number.

Thus, suppose if a rectangle is dilated, and its sides were of length = L and width = W, then its dilated version would be having length = Ln, and width = Wn where n is the factor of scaling.

Thus, we get:

$n = \dfrac{Ln}{L} = \dfrac{Wn}{W}\\\\or\\\\\text{Ratios of corresponding dilated sides are equal}$

For the given cases, checking all the given pairs:

Case 1: A rectangle with length 4 and width 2. A rectangle with length 8 and width 4.

Getting length to length and width to width ratio:

$\dfrac{L_1}{L_2} = \dfrac{4}{8} = \dfrac{1}{2}\\\\\dfrac{W_1}{W_2} = \dfrac{2}{4} = \dfrac{1}{2}\\\\\\Thus, \dfrac{L_1}{L_2} = \dfrac{W_1}{W_2}$

Pair given are dilated version of each other.

Case 2: A rectangle with length 4 and width 2. A rectangle with length 6 and width 3.

Getting length to length and width to width ratio:

$\dfrac{L_1}{L_2} = \dfrac{4}{6} = \dfrac{2}{3}\\\\\dfrac{W_1}{W_2} = \dfrac{2}{3} \\\\\\Thus, \dfrac{L_1}{L_2} = \dfrac{W_1}{W_2}$

Pair given are dilated version of each other.

Case 3: A rectangle with length 4 and width 3. A rectangle with length 3 and width 2.

Getting length to length and width to width ratio:

$\dfrac{L_1}{L_2} = \dfrac{4}{3}\\\\\dfrac{W_1}{W_2} = \dfrac{3}{2} \\\\\\Thus, \dfrac{L_1}{L_2} \neq \dfrac{W_1}{W_2}$

Pair given are not dilated version of each other.

Case 4: A rectangle with length 4 and width 3. A rectangle with length 2 and width 1.5.

Getting length to length and width to width ratio:

$\dfrac{L_1}{L_2} = \dfrac{4}{2} = \dfrac{2}{1}\\\\\dfrac{W_1}{W_2} = \dfrac{3}{1.5} = \dfrac{2}{1}\\\\\\Thus, \dfrac{L_1}{L_2} = \dfrac{W_1}{W_2}$

Pair given are dilated version of each other.

Thus, the pair of rectangle that will prove Carly's statement incorrect is: Option 3: A rectangle with length 4 and width 3. A rectangle with length 3 and width 2.

Learn more about dilation here:

brainly.com/question/3266920

5 0

2 years ago

Read 2 more answers