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lozanna [386]
2 years ago
15

A physical fitness association is including the mile run in its secondary school fitness test. The time for this event for boys

in secondary school is known to possess a normal distribution with a mean of 440 seconds and a standard deviation of 40 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.A) 0.0107 B) 0.9893 C) 0.5107 D) 0.4893
Mathematics
1 answer:
Ierofanga [76]2 years ago
6 0

Answer:

A) 0.0107

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 440 seconds and a standard deviation of 40 seconds.

This means that \mu = 440, \sigma = 40

Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.

This is the p-value of Z when X = 348. So

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

Z = \frac{348 - 440}{40}

Z = -2.3

Z = -2.3 has a p-value of 0.0107, and thus, the correct answer is given by option A.

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Solve the system by graphing. Write the solution as an ordered pair. y = 1/3x + 2 y = –x – 2
Dima020 [189]
ANSWER

The solution is where the two graphs intersect, which is
(-3,1).

EXPLANATION

The given system of equations are
y = \frac{1}{3} x + 2
and

y = - x - 2

We need to graph the two equations.

Let us graph

y = \frac{1}{3} x + 2
first.

We need at least two points.

You can choose any appropriate value for x and solve for y. Choosing zero makes our working easier. So let us plot the intercepts.

When
x = 0

\Rightarrow \: y = \frac{1}{3} (0) + 2

\Rightarrow y = 0 + 2

\Rightarrow y = 2

So this gives us the ordered pair,

(0,2)

When
y = 0
we get,

0= \frac{1}{3} x + 2

\Rightarrow \: - 2 = \frac{1}{3} x

\Rightarrow \: - 2 \times 3= x

\Rightarrow \: -6= x
This also gives the ordered pair

(-6,0).

We plot these two points and draw a straight line through them to obtain the blue graph in the attachment.

For the second line

y = - x - 2

We again find the intercepts and plot them.

When
x = 0

y = - 0 - 2

\Rightarrow \: y = - 2

This gives the ordered pair

(0,-2)

Also, when
y = 0

then we have,

0 = - x - 2

2 = - x

x = - 2

Then we again have the ordered pair,

(-2,0)

We plot these two points on the same graph sheet to obtain the red graph above.

The intersection of the two lines is
(-3,1)


You will get good grades so don't worry much.

4 0
3 years ago
Name a chord for the circle.​
kobusy [5.1K]

Answer:

several choices (see below and refer to attached))

Step-by-step explanation:

A chord is a straight line that have BOTH endpoints on the circumference

in our case, we have several chords, just pick one

RN

RP

MN

QN

PQ

8 0
3 years ago
Which real world image represents a negative slope?
salantis [7]

Answer:

The soccer goal

Step-by-step explanation:

A negative slope always points down.

4 0
3 years ago
.In the Star Wars franchise, Yoda stands at only 66 centimeters tall. Suppose you want to see whether or not hobbits from the Lo
r-ruslan [8.4K]

Answer:

We conclude that the average height of hobbit is taller than Yoda.

Step-by-step explanation:

We are given that in the Star Wars franchise, Yoda stands at only 66 centimetres tall.

From a sample of 7 hobbits, you find their mean height \bar X = 80 cm with standard deviation s = 10.8 cm.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average height of hobbit.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 66 cm      {means that the average height of hobbit is shorter than or equal to Yoda}

Alternate Hypothesis, H_A : \mu > 66 cm      {means that the average height of hobbit is taller than Yoda}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                     T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}  ~ t_n_-_1

where, \bar X = sample mean height = 80 cm

             s = sample standard deviation = 10.8 cm

             n = sample of hobbits = 7

So, <u><em>test statistics</em></u>  =  \frac{80-66}{\frac{10.8}{\sqrt{7}}}  ~ t_6

                              =  3.429

The value of t test statistics is 3.429.

<em>Now, at 0.01 significance level the t table gives critical value of 3.143 for right-tailed test. Since our test statistics is more than the critical value of t as 3.429 > 3.143, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

<em />

Therefore, we conclude that the average height of hobbit is taller than Yoda.

3 0
3 years ago
Use properties of addition and subtraction to evaluate the expression.<br><br> −28−47−12
Brrunno [24]
The answer would be -87 
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2 years ago
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