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Yakvenalex [24]
3 years ago
11

The weight of fish in Lake Paradise follows a normal distribution with mean of 7.5 lbs and standard deviation of 2.5 lbs. a. Wha

t proportion of fish are between 9lbs and 12lbs? Give your answer to 3 decimal places. b. Alex boasts that he once caught a fish that was just big enough to be in the top 3% of of the fish population. How much did his fish weigh? Give your answer to 2 decimal places. c. If one catches a fish from the bottom 20% of the population, the fish must be returned to the lake. What is the weight of the smallest fish that one can keep? Give your answer to 2 decimal places.
Mathematics
1 answer:
hoa [83]3 years ago
7 0

Answer:

oooooh

Step-by-step explanation:

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In a random sample of cars driven at low altitudes, of them exceeded a standard of grams of particulate pollution per gallon of
Orlov [11]

Complete question is;

In a random sample of 370 cars driven at low altitudes, 43 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 80 cars driven at high altitudes, 23 of them exceeded the standard. Can you conclude that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard at an level of significance? Group of answer choices

Answer:

Yes we can conclude that there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard (P-value = 0.00005).

Step-by-step explanation:

This is a hypothesis test for the difference between the proportions.

The claim is that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

Then, the null and alternative hypothesis are:

H0 ; π1 - π2 = 0

H1 ; π1 - π2 < 0

The significance level would be established in 0.01.

The random sample 1 (low altitudes), of size n1 = 370 has a proportion of;

p1 = x1/n1

p1 = 43/370

p1 = 0.116

The random sample 2 (high altitudes), of size n2 = 80 has a proportion of;

p2 = x2/n2

p2 = 23/80

p2 = 0.288

The difference between proportions is pd = (p1-p2);

pd = p1 - p2 = 0.116 - 0.288

pd = -0.171

The pooled proportion, we need to calculate the standard error, is:

p = (x1 + x2)/(n1 + n2)

p = (43 + 23)/(370 + 80)

p = 66/450

p = 0.147

The estimated standard error of the difference between means is computed using the formula:

S_(p1-p2) = √[((p(1 - p)/n1) + ((p(1 - p)/n2)]

1 - p = 1 - 0.147 = 0.853

Thus;

S_(p1-p2) = √[((0.147 × 0.853)/370) + ((0.147 × 0.853)/80)]

S_(p1-p2) = 0.044

Now, we can use the formula for z-statistics as;

z = (pd - (π1 - π2))/S_(p1-p2)

z = (-0.171 - 0)/0.044

z = -3.89

Using z-distribution table, we have the p-value = 0.00005

Since the P-value of (0.00005) is smaller than the significance level (0.01), then the effect is significant.

We conclude that The null hypothesis is rejected.

Thus, there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

6 0
3 years ago
Does anyone Know The answer to this Graph!
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Alright, so i’ll make a graph for you with the points.... just look at the image below. i hope you give me brainliest!!! i’m trying hard!

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4 0
3 years ago
The diagram shows a rectangular garden path.
marta [7]

Answer:

Step-by-step explanation:12

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3 years ago
Is the factored form of 4k^2-28k+49
oksano4ka [1.4K]
It should be (2k - 7)^2
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The play director spent 190190190190 hours preparing for a play. That time included attending 35353535 rehearsals that took vary
Genrish500 [490]

Answer:

The equation above represents the total time the play director spent preparing for a play.

Step-by-step explanation:

The time spent by the play director for preparing for a play is, 190 hours.

Of these 190 hours, the director spent varying amounts of time attending 35 rehearsals for the play.

Let the varying amounts of time be denoted by, <em>x</em>.

The director also spent 3/4th of an hour, i.e. 45 minutes, on other responsibilities related to the play.

The equation provided is:

35x+\frac{3}{4}=190

The equation above represents the total time the play director spent preparing for a play.

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4 years ago
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