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2 years ago

8

You randomly select one card from a 52-card deckFind the probability of selecting a black ace or a red nine

1 answer:

rosijanka [135]2 years ago

6 0

Answer:

2/52 + 2/52 = 4/52

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By how much is the product of 9/5 and 8 1/4 greater than 5

Oliga [24]

Given:

The product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ greater than 5.

To find:

By how much the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ greater than 5.

Solution:

First we need to find the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ .

$\text{Product}=\dfrac{9}{5}\times 8\dfrac{1}{4}$

$\text{Product}=\dfrac{9}{5}\times \dfrac{8(4)+1}{4}$

$\text{Product}=\dfrac{9}{5}\times \dfrac{33}{4}$

$\text{Product}=14.85$

Now subtract 5 from the product.

$14.85-5=9.85$

Therefore, the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ is 9.85 greater than 5.

3 0

2 years ago

How much greater is this week’s median temperature than last week’s median temperature?

Scrat [10]

Answer:

5 degrees

Step-by-step explanation:

5 0

2 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLIEST AND POINTS

ZanzabumX [31]

Answer:

bcdf

Step-by-step explanation:

here ya go

4 0

2 years ago

HELP ASAPPPP NOWWW PLEASEEE

pickupchik [31]

X=0 because when you have something to the power of zero your answer is 1.
x=0 since 1 to the power of zero equals 1.

hope im not too late

5 0

2 years ago

Read 2 more answers