Please answer please

Suppose that two cards are randomly selected from a standard 52-card deck. (a) what is the probability that the first card is

eduard

You draw a queen on the first draw and one on the second4/52 ∗ 3/51 = 12/2652 = 1/22 
You draw something other than a queen on the first draw and a queen on the second
48/52 ∗ 4/51 = 192/2652=16/221

So in total, the chance is:

1/221 + 16/221 = 17/221=
4/52

8 0

4 years ago

Helpppppppppppp plzzzzzzzzzzzz!

r-ruslan [8.4K]

For the first one:
2x + 2 = 3x - 52
2x + 2 + 52 = 3x - 52 + 52
2x + 54 = 3x
2x - 2x + 54 = 3x -2x
54 = x
For the second:
124 + 2x + 4 = 180
124 + 4 + 2x = 180
128 + 2x = 180
128 - 128 + 2x = 180 - 128
2x = 52
2x/2 = 52/2
x = 26

6 0

4 years ago

A=2πr solve for π‍‍‍

pishuonlain [190]

$A=2\pi r\\\\2\pi r=A\ \ \ \ |divide\ both\ sides\ by\ 2r\\\\\boxed{\pi=\frac{A}{2r}}$

6 0

3 years ago

Plz help!!!! I really need the help

Alexeev081 [22]

can you tell me more?

like what you were given

6 0

3 years ago

A publisher reports that 65% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is

muminat

Answer:

$z=\frac{0.60 -0.65}{\sqrt{\frac{0.65(1-0.65)}{340}}}=-1.933$

The p value for this case can be calculated with this probability:

$p_v =2*P(z$

For this case is we use a significance level of 5% we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is different from 0.65 or 65%. We need to be careful since if we use a value higher than 65 for the significance the result would change

Step-by-step explanation:

Information given

n=340 represent the random sample taken

$\hat p=0.60$ estimated proportion of readers owned a laptop

$p_o=0.65$ is the value that we want to test

z would represent the statistic

$p_v{/tex} represent the p valueHypothesis to testWe want to check if the true proportion of readers owned a laptop if different from 0.65
Null hypothesis:[tex]p=0.65$

Alternative hypothesis: $p \neq 0.65$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.60 -0.65}{\sqrt{\frac{0.65(1-0.65)}{340}}}=-1.933$

The p value for this case can be calculated with this probability:

$p_v =2*P(z$

For this case is we use a significance level of 5% we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is different from 0.65 or 65%. We need to be careful since if we use a value higher than 65 for the significance the result would change

5 0

4 years ago