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Pani-rosa [81]
3 years ago
10

Lake Lopez had a water depth at its dam of 515 inches at the beginning of the year. During the rainy season, the water depth inc

reased by 82 inches. Evaporation during a hot summer decreased the depth by 119 inches. What was the water depth at the end of the summer?
Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
7 0

Beginning of the year= 515 inches

-

<u>Rain increases the water depth:</u>

After rainy season = 515 + 82 = 597 inches

-

<u>Evaporation decreases the water depth:</u>

After Hot Summer = 597 - 119 = 478 inches

-

Answer: 478 inches

guapka [62]3 years ago
4 0

This question states that Lake Lopez began at a depth of 515 inches, then increased by 82 inches, then decreased by 119 inches. To model this, we will use addition for the increase and subtraction for the decrease; this will signify a positive/negative change in the depth. This can be modeled by the expression below:

515 + 82 - 119

597 - 119

478

Therefore, the water depth at the end of the summer was 478 inches.

Hope this helps!

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2 years ago
Compute​ P(x) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate thi
stich3 [128]

Answer:

The answers to the questions are;

A) P(X) where x = 12 is equal to 3.55 ×10⁻²

B) P(x) using the normal distribution is given by the probability distribution function and is equal to 3.453 ×10⁻²

C) The calculated probabilities of P(x=12) using the binomial probability formula and the probability distribution function differ by 9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     a. because np(1-p) %u2265 ⇒ np(1-p) ≥ 10

E) c. the value of fi represents the expected proportion of observation less than or equal to the ith data value.

Step-by-step explanation:

To solve the question, we note that

n = 58

p = 0.3

x = 12

Here we have n·p = 17.4 and n·q = 40.6 both ≥ 5 that is the normal distribution can be used to estimate the probability

A) We are to use the binomial probability formula to find P(X)

Where x = 12, we have P(12) = ₅₈C₁₂×0.3¹²×0.7⁴⁶

= 891794789340×0.000000531441×7.49×10⁻⁸ = 3.55 ×10⁻²

B) Using the standard distribution table we have

the z score for x = 12 given as z = \frac{x-\mu}{\sigma} = -1.55

From the normal distribution table, we have the probability that value is below 12 = .06057 while the normal probability distribution function which gives the probability of a number being 12 is given by

\frac{1}{\sigma\sqrt{2 \pi } } e^{\frac{-(x-\mu)^2}{2\sigma^2} }

where:

σ = Standard deviation = \sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999

μ = Sample mean = n·p = 58×0.3 =17.4

Therefore the probability density function is \frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}

C)  The probabilities differ by 3.55 ×10⁻² -  3.453 ×10⁻² =  9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     n·p and n·p·(n-p) ≥ 5

E)   f_i represents the area under the curve towards left of the ith data observed in a normally distributed population.

Therefore, the value of fi represents the expected proportion of observation less than or equal to the ith data value  

5 0
2 years ago
Please answer quick
Musya8 [376]

Answer:

x can be transformed into -1/2x

-1/2x / -1/2=x

1/-1/2=-2

x>-2

Step-by-step explanation:

Tell me if it’s too confusing or if you got it hope it help sorry if it didn’t

3 0
3 years ago
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