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Katyanochek1 [597]
3 years ago
7

What’s this need help asap !

Mathematics
1 answer:
Vsevolod [243]3 years ago
5 0
I need points and i’m using this bc you didn’t even give an equation
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The annual snowfall in city Ais 43.9 inches. This is 13.5 inches more than four times the snowfall in city B. Find the annual sn
Jlenok [28]

Answer:

24.475

Step-by-step explanation:

1st Divide: 43.9÷4=10.975

2nd Add 13.5: 10.975+13.5=24.475

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Hope this helps you out :)

7 0
3 years ago
Write an equality with the solution x <12
Tom [10]
Lol i learned this last chapter. ok here is one 2x + 3 < 5
4 0
3 years ago
The probability of flu symptoms for a person not receiving any treatment is 0.038. In a clinical trial of a common drug used to
alexgriva [62]

Answer:

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 1164, p = 0.038

So

\mu = E(X) = np = 1164*0.038 = 44.232

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 6.5231

Estimate the probability that at least 47 people experience flu symptoms.

Using continuity correction, this is P(X \geq 47 - 0.5) = P(X \geq 46.5), which is 1 subtracted by the pvalue of Z when X = 46.5.

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

Z = \frac{46.5 - 44.232}{6.5231}

Z = 0.35

Z = 0.35 has a 0.6368

1 - 0.6368 = 0.3632

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

6 0
3 years ago
What is the value of 0.1561 rounded to the nearest tenth?
Ainat [17]

We have to round the value of 0.1561 to the nearest tenth.

The number after decimal is the number at tenth place. Consider the number to the right of the tenths place and use the number to determine if you will round up or stay the same. Notice that the number to the right of tenth place is more than or equal to 5 or less than 5. If that number is greater than or equal to 5, then the number will round up but if that number is less than 5, then the number will not round up. It will remain same.

Let us consider the given number 0.1561

The number at tenths place is 1

The number after the tenths place is 5 (which is either greater than or equal to 5)

So, the number will round up to 0.2

8 0
3 years ago
Read 2 more answers
How do I find the slope, domain, range and y intercept of y=-2x
Natasha2012 [34]

Slope is defined as the rate of change, in this case the change in y with respect to x. In standard form for a binomial linear equation, y = mx + b. The constants m and b are the slope and intercept, respectively. For your equation, the slope is -2 and the intercept is 0. The domain (possible values of x) for this graph is (-∞, ∞) and the range (possible values of y) is also (-∞, ∞).

5 0
3 years ago
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