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Gre4nikov [31]
3 years ago
15

Please answer please

Mathematics
1 answer:
love history [14]3 years ago
7 0

Answer:

43.5 cm²

Step-by-step explanation:

½ × 12 × 8 × sin(65)

= 43.50277378 cm²

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Suppose that two cards are randomly selected from a standard​ 52-card deck. ​(a) what is the probability that the first card is
eduard
<span><span>You draw a queen on the first draw and one on the second<span><span><span>4/52 </span>∗ <span>3/51 </span>= <span>12/2652 </span>= <span>1/22 </span></span></span></span><span>
You draw something other than a queen on the first draw and a queen on the second<span><span><span>
48/52 </span>∗ <span>4/51 </span>= <span>192/2652</span>=<span>16/221</span></span></span></span></span>

So in total, the chance is:

<span><span><span>1/221 </span>+ <span>16/221 </span>= <span>17/221</span>=
<span>4/52</span></span></span>

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&#10;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
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