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Lady bird [3.3K]

4 years ago

="latex-formula">

Mathematics

2 answers:

7nadin3 [17]4 years ago

7 0

Answer:

Step-by-step explanation:

If you have 2 pizzas and 2 more pizzas you have 4 pizzas.

Ganezh [65]4 years ago

3 0

Answer:

Step-by-step explanation:

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Five thousand tickets are sold at $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded a

Elis [28]

Answer:

The expected value of this raffle if you buy 1 ticket is $0.41.

Step-by-step explanation:

The expected value of the raffle if we buy one ticket is the sum of the prizes multiplied by each of its probabilities.

This can be written as:

$E(X)=\sum p_iX_i$

For example, the first prize is $800 and we have only 1 prize, that divided by the number of tickets gives us a probability of 1/5000.

If we do this with all the prizes, we can calculate the expected value of a ticket.

$E(X)=\sum p_iX_i\\\\\\E(X)=\dfrac{1\cdot800+3\cdot200+5\cdot50+20\cdot20}{5000}\\\\\\E(X)=\dfrac{800+600+250+400}{5000}=\dfrac{2050}{5000}=0.41$

7 0

3 years ago

Solve by factoring<br> 3v^2-10v+3=

jolli1 [7]

$3v^2-10v+3=0\\\\\implies 3v^2 - 9v-v +3=0\\ \\\implies 3v(v-3)-(v-3)=0\\\\\implies (v-3)(3v-1)=0\\\\\implies v = 3,~~v=\dfrac 13$

8 0

3 years ago

Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx

Rudiy27

If we use the substitution $u = \tan x$ , then $du = \sec^2 {x}\ dx$ . If you try substituting just $u$ and $du$ into the integrand, though, you'll notice that there's a $\sec^2x$ left over that we have to deal with.

To get rid of this problem, use the identity $\tan^2 x + 1 = \sec^2 x$ and substitute in the left side of the identity for the extra $\sec^2x$ , as shown:

$\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx$
$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$

From there, we can substitute in $u$ and $du$ , and then evaluate:

$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du$
$= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}$
$= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})$
$= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}$

8 0

3 years ago

Use the numbers 3 and 4 to make a fraction that is greater than 1 and a fraction less than 1. Explain how you made your fraction

Lana71 [14]

Answer:

4/3

Step-by-step explanation:

put one number on the top and one on the bottom. trial and error...

3 0

3 years ago

Read 2 more answers

-3[10-(3-8)+4-2²-3(4-9)]

NISA [10]

−3(10−(3−8)+4−2²-−3(4−9))

=−3(10−(−5)+4−2²-−3(4−9))

=−3(15+4−2²-−3(4−9))

=−3(19−2²-−3(4−9))

=−3(19−4−3(4−9))

=−3(15−3(4−9))

=−3(15−(3)(−5))

=−3(15−(−15))

=(−3)(30)

=−90

4 0

3 years ago