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3 years ago

PLEASE HELP ME WITH THIS QUESTION.

Mathematics

1 answer:

erik [133]3 years ago

3 0

The left one can be 1, 2, 3, or 4 .
The right one can be 2, 4, 6, or 8.

Their sum can be (2, 4, 6, or 8) plus (1, 2, 3, or 4) .

The smallest possible sum is (2) + (1) = 3 .
The largest possible sum is (8) + (4) = 12 .
And it could be any number in between those.
Select the choice that says this.

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A woman has 7 keys. only one of them will open her door. when seh arrives at the door there is no sufficient light for her to fi

Gala2k [10]

Since she has seven key the probability that the first one will open the door is
$\frac{1}{7}$
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$\frac{1}{3}$
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hope this helps

7 0

3 years ago

Change the decimal into a %:0.7

Tomtit [17]

Simply multiply 0.7 by 100. 0.7 x 100= 70%. The answer is 70%

8 0

3 years ago

Read 2 more answers

Please Help

Sunny_sXe [5.5K]

Joel read the greatest number of pages in a day.

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3 years ago

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WILL GIVE BRAINLIEST FAST!! Will Give Brainliest if you show all the steps. 8x - 6 / 2 = x + 12

Naddik [55]

Answer: 15/7 2 1/7

Step-by-step explanation: First divide 8x-3=x+12 Then add 3 to both sides.

8x=x+15 Subtract x from both sides. 7x=15 Then divide again so 7/7 and 15/7. If you simplify it become 2 1/7

4 0

3 years ago

Read 2 more answers

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Vikentia [17]

Answer:

$\= x = 18.5$ , $\sigma = 5.15$

$15.505 < \mu < 21.495$

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

8 0

3 years ago