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Lubov Fominskaja [6]
2 years ago
11

When Gabriella runs the 400 meter dash, her finishing times are normally distributed with a mean of 79 seconds and a standard de

viation of 0.5 seconds. Using the empirical rule, determine the interval of times that represents the middle 95% of her finishing times in the 400 meter race.
Mathematics
1 answer:
kupik [55]2 years ago
4 0

Answer:

(78, 80)

Step-by-step explanation:

Given a normal distribution :

Mean = 79 seconds

Standard deviation = 0.5 seconds

Using the empirical rule :

95% of her finishing times :

According to the empirical rule, 95% represents 2 standard deviations from the mean that is ± 2 standard deviations from the mean score or value.

Hence ;

The interval will be defined as ;

Mean ± 2(standard deviation)

79 ± 2(0.5)

79 ± 1

(79 - 1) ; (79 + 1)

78 ; 80

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nexus9112 [7]

Answer:

  • 3*10^{-6} or 0.000003

Given expression:

  • (6.3*10^{-2}) : (2.1*10^4)

Simplify it in steps below:

  • (6.3*10^{-2}) : (2.1*10^4) =
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<u>Used properties:</u>

\cfrac{ab}{cd}=\cfrac{a}{c}*\cfrac{b}{d}

a^b:a^c=a^{b-c}

a^{-b}=1/a^b

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A = P + I where

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I (interest) = $ 825.22

Third Answer: A = $ 7,735.55

A = P + I where

P (principal) = $ 7,000.00

I (interest) = $ 735.5

Fourth Answer: A = $ 7,735.55

A = P + I where

P (principal) = $ 7,000.00

I (interest) = $ 735.55

Step-by-step explanation:

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1300 error interval to 2dp
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Step-by-step explanation:

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Solve for x<br><br> A. 9<br> B. 10<br> C. 42<br> D. 12
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Option D, 12

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X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

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