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AlekseyPX
2 years ago
8

Jamie has a deck of 60 sports cards, of which some are baseball and summer football cards. Jamie pulls out a card randomly from

the deck, record it's type, and replaces it in the deck. Jamie has already recorded 6 baseball cards and 9 football cards based on these data what is most likely the number of baseball cards in the deck?
Mathematics
1 answer:
mixer [17]2 years ago
7 0

Answer:

3 baseball + 9 football  9+3 = 12

3 out of 12 were base ball

3/12=0.25

60*0.25 = 15

so there would be 15 baseball cards

Step-by-step explanation:

You might be interested in
354 by rounding to the nearest hundred.​
lara [203]

Answer:

400

Step-by-step explanation:

354>=350 and thus is rounded upwards to 400.

4 0
3 years ago
Read 2 more answers
A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
2 years ago
I roll a four-sided die. The possible outcomes are 1, 2, 3, or 4, depending on the number ot spots on the side of the die that i
Daniel [21]

Answer:

The sample space

Step-by-step explanation:

By definition, the sample space is the set composed by all the possible outcomes of your random variable. In this case, the random variable is 4 sided dice, so the sample space is

\Omega=\{1,2,3,4\}

8 0
3 years ago
In 2002, the population of a town was approximately 34,050. In 2012,
mrs_skeptik [129]
It’s D
hope that helps
5 0
2 years ago
Walter is practicing his high jumps for the track team. His first jump measured 5 feet, 9 inches. His second jump measured 3 fee
tia_tia [17]

Answer: 1.834 ft

Step-by-step explanation:

Given

Walter's first Jump is 5 ft, 9 in.

His second Jump is 3 ft, 11 in.

We know,

1\ ft=12\ in

Converting inches to feet

First Jump in feet

\Rightarrow 5\ ft+\frac{9}{12}\ ft=5.75\ ft

Second jump

\Rightarrow 3\ ft+\frac{11}{12}\ ft=3.916\ ft

The difference in the measurement of two Jumps is

\Rightarrow 5.75-3.916=1.834\ ft

So, Walter first Jump is 1.834 ft higher than the second Jump

3 0
3 years ago
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