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Savatey [412]
3 years ago
5

PLEASE HELP IM STUCK ON THIS QUESTION

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
3 0
THe answer is the second option because you have to use the pythogorean theorm to fing the 3rd side and multiply your answer by 2 to get the value of x.
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I need help on how to get the answer too number 27.
KiRa [710]
A^2 + b^2 = c^2
3^2 + 8^2 = c^2
9 + 64 = c^2
73 =c^2
take the square root of each side
c= 8.544

3 0
2 years ago
Is this correct or no <br><br> Please help me
k0ka [10]

Answer:

I got y = 10

or

y = 24

or the answer is y = -5x since the 4 stays the same but the -5x changes

I feel like you have to replace with x with the numbers

Step-by-step explanation:

3 0
3 years ago
After the drama club sold 100 tickets to a show, it had $300 in profit. After the next show, it had sold a total of 200 tickets
ZanzabumX [31]

Answer:

B. y-300=4(x-100)

Step-by-step explanation:

4 0
3 years ago
DP
irinina [24]

The probability of drawing a two and another 2 is 1/72

<h3>What is probability?</h3>

Probability is the likelihood or chance that an event will occur

  • Since the total number in the board game is 9, hence the probability of drawing a 2 will be 1/9

  • The probability of drawing the second 2 will be 1/8

  • Pr(drawing a 2, then drawing another 2) = 1/9 * 1/8

Pr(drawing a 2, then drawing another 2) = 1/72

Hence the probability of drawing a two and another 2 is 1/72

learn more on probability here: brainly.com/question/24756209

4 0
1 year ago
insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of
ss7ja [257]

Answer: (13.22\%,\ 19.34\%)

Step-by-step explanation:

Given : Sample space : n= 559

Sample proportion : \hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628

=16.28\%

Significance level : \alpha= 1-0.95=0.05

Critical value : z_{\alpha/2}=1.96

Confidence level for population proportion:-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.1628\pm (1.96)\sqrt{\dfrac{0.1628(1-0.1628)}{559}}\\\\=0.1628\pm0.030604971367\\\\\approx 0.1628\pm0.0306=(0.1322,\ 0.1934)=(13.22\%,\ 19.34\%)

Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= (13.22\%,\ 19.34\%)

7 0
3 years ago
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