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NISA [10]
2 years ago
8

Write a problem about your favorite television show that uses the equation x + 8 = 30.

Mathematics
1 answer:
rosijanka [135]2 years ago
3 0

Answer:

Step-by-step explan

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What is 5x=-2y+6 in slope-intercept form?
Gemiola [76]

Answer:

y=-5/2x+3

Step-by-step explanation:

8 0
3 years ago
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Let z denote a variable that has a standard normal distribution. Determine the value z* to satisfy the following conditions. (Ro
Gre4nikov [31]

Answer:

a) z* = -1.97

b) z* = -2.33

c) z* = -1.65

d) z* = 2.04

e) z* = 2.33

f) z* = -1.25.

Step-by-step explanation:

Z-score:

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.

a. P(z < z*) = 0.0244

We have to look at the ztable, and find z which has a pvalue of 0.0244. So it is z* = -1.97

b. P(z < z*) = 0.0098

We have to look at the ztable, and find z which has a pvalue of 0.0098. So it is z* = -2.33

c. P(z < z*) = 0.0496

We have to look at the ztable, and find z which has a pvalue of 0.0496. So it is z* = -1.65

d. P(z > z*) = 0.0204

We have to look at the ztable, and find z which has a pvalue of 1 - 0.0204 = 0.9796. So z* = 2.04

e. P(z > z*) = 0.0098

We have to look at the ztable, and find z which has a pvalue of 1 - 0.0098 = 0.9902. So z* = 2.33

(f) P(z > z* or z < -z*) = 0.201

This is z which has a pvalue of 0.201/2 = 0.1055. So it is z* = -1.25.

4 0
3 years ago
Read 2 more answers
Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1/ 8. If someone arrives
Elis [28]

Answer:

a) P=0.535

b) P=0.204

c) P=0.286

Step-by-step explanation:

The exponential distribution is expressed as

F(x>t)=e^{-\lambda t}

In this example, λ=1/8=0.125 min⁻¹.

a) The probability of having to wait more than 5 minutes

F(x>5)=e^{-0.125*5}=e^{-0.625}=0.535

b) The probability of having to wait between 10 and 20 minutes

F(1020)=e^{-0.125*10}-e^{-0.125*20}\\\\F(10

c) The exponential distribution is memory-less, so it is independent of past events.

If you have waited 5 minutes, the probability of waiting more than 15 minutes in total is the same as the probability of waiting 15-5=10 minutes.

F(x>15|t^*=5)=F(x>15)/F(x>5)=\frac{e^{-0.125*15}}{e^{-0.125*5}}=e^{-0.125*(15-5)}\\\\ F(x>15|t^*=5)=e^{-0.125*(10)}=F(x>10)=0.286

5 0
3 years ago
Help please, I’m not sure
JulijaS [17]
\frac{ \frac{7}{24} }{ \frac{35}{48} } = \frac{2}{5}
4 0
3 years ago
The demand function for a particular product is given by D(x)=−0.75x2+5x+50‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
galben [10]

Step-by-step explanation:

hi nice to meet you my name is anna

6 0
3 years ago
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