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3 years ago

The football team has 50 jerseys. There are 15 medium-sized jerseys. What percent of the jerseys are medium sized jerseys?

Mathematics

1 answer:

icang [17]3 years ago

8 0

15/50 = 3/30 = 1/10 1/10 = 10% That is about as simple as I could put it.

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In 2012 your car was worth $10,000. In 2014 your car was worth $8,800. Suppose the value of your car decreased at a constant rat

Sonbull [250]

Answer:

g(t) = 10000(0.938)^t

Step-by-step explanation:

Given data:

car worth is $10,000 in 2012

car worth is $8000 in 2014

let linear function is given as

P(t) = at + b

which denote the value of car in year t

take t =0 for year 2012

at t =0, 10,000 = 0 + b

we get b = 10,000

take t =2 for year 2014

at t =2, P(2) = 2a + b

8800 = 2a + 10,000

a = - 600

Thus the price of car at year t after 2012 is given as p(t) = -600t + 10000

let the exponential function $p(t) =ab^2$ where t denote t = 0 at 2012

putting t = 0 P(0) = 10,000 we get 10,000 = ab^0

a = 10,000

putting t = 2 p = 8800

$8800 =ab^2$

$b^2 = \frac{8800}{10000}$

b = 0.938

g(t) = 10000(0.938)^t

3 0

3 years ago

Peter wallpapered a wall that was 9 feet wide and 8 feet high. He had 27 square feet of wallpaper left over. How many square fee

ivanzaharov [21]

Answer:

99 square feet

Step-by-step explanation:

Because 8 times 9=72+27=99

7 0

2 years ago

Read 2 more answers

If the model boat is 8 cm long how long is the full size boat?

sertanlavr [38]

Answer:

1 inch equals 8 feet

Step-by-step explanation:

7 0

2 years ago

King kyle orders the construction of his future mausoleum ( a fancy tomb) so that people never forget his awesomeness. His archi

BartSMP [9]

Answer:

okay so,

so do a 16 by 14 rectangle it should be good

sry if its wrong i tried but i hope it helped ;0

5 0

2 years ago

Read 2 more answers

Quadrant:

NISA [10]

<h2>✒️Area Between Curves</h2>

$\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}$

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5 0

2 years ago