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Vinil7 [7]
3 years ago
5

Please help and show work​

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
4 0

Answer:

50 : 3

Step-by-step explanation:

Step 1: Start with the basic ratio

Adult Panda : Baby Panda

200 : 12

Step 2: Simplify

200 : 12

100 : 6

50 : 3

Step 3: Therefore Statement

Therefore the ratio of an adult panda to a baby panda's wieight is 50 : 3

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B) sunlight travels through space - there is nothing to bounce off of until it hits an atmophere
3 0
3 years ago
Read 2 more answers
10
timama [110]

Answer:

25π units^2

Step-by-step explanation:

The shape is a quarter circle with radius 10 units.

area of circle = πr^2

area of a quarter circle = (1/4)π(10 units)^2

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5 0
3 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
Evaluate the following expression: (-5)(6) + (-6) ÷ (-3). Remember to show your work
Vinil7 [7]

Answer:

-28

Step-by-step explanation:

(-5)(6) + (-6) ÷ (-3) =

-30 + (-6) ÷ (-3) =

-30 + 2 =

-28

<em>Hope that helps! :)</em>

<em></em>

<em>-Aphrodite</em>

7 0
3 years ago
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