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strojnjashka [21]
3 years ago
13

Point P on the number line shows Lara's score after the first round of a quiz: A number line is shown from negative 8 to 0 to po

sitive 8. There are increments of 1 on either side of the number line. The even numbers are labeled on either side of the number line. Point P is shown on 3. In round two, she lost 4 points. Which expression shows how many total points she has at the end of round two? (5 points) Group of answer choices 3 + 1 = −4, because −4 is 1 unit to the left of 3 3 + (−1) = −4, because −4 is 1 unit to the left of 3 3 + 4 = −1, because −1 is 4 units to the left of 3 3 + (−4) = −1, because −1 is 4 units to the left of 3
Mathematics
2 answers:
Aloiza [94]3 years ago
6 0

Answer: its D

Step-by-step explanation:

Luda [366]3 years ago
3 0

Answer:

The correct group of answer is D) 3 + (−4) = −1, because −1 is 4 units to the left of 3

Step-by-step explanation:

Consider the provided information.

Point P on the number line shows Lara's score after the first round of a quiz:

Point P is shown on 3.

That means after the first round her total was 3.

In round two, she lost 4 points.

Therefore, she score −4 points in 2nd round

To find the total points at the end of round 2, we need to add -4 to 3.

3+(−4)=−1

Here −1 is 4 units to the left of 3.

Hence, the correct group of answer is D) 3 + (−4) = −1, because −1 is 4 units to the left of 3

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Construct the confidence interval for the population standard deviation for the given values. Round your answers to one decimal
Alex777 [14]

Answer:

The correct answer is "2.633< \sigma < 4.480".

Step-by-step explanation:

Given:

n = 21

s = 3.3

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now,

df = n-1

    =20

⇒ x^2_{\frac{\alpha}{2}, n-1 } = x^2_{\frac{0.9}{2}, 21-1 }

                  = 31.410

⇒ x^2_{1-\frac{\alpha}{2}, n-1 } = 10.851

hence,

The 90% Confidence interval will be:

= \sqrt{\frac{(n-1)s^2}{x^2_{\frac{\alpha}{2}, n-1 }} } < \sigma < \sqrt{\frac{(n-1)s^2}{x^2_{1-\frac{\alpha}{2}, n-1 }}

= \sqrt{\frac{(21-1)3.3^2}{31.410} } < \sigma < \sqrt{\frac{(21.1)3.3^2}{10.851} }

= \sqrt{\frac{20\times 3.3^2}{31.410} } < \sigma < \sqrt{\frac{20\times 3.3^2}{10.851} }

= 2.633< \sigma < 4.480

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