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Varvara68 [4.7K]

2 years ago

4

A student saves money to buy a laptop. The student already has 307.50 saved and puts aside $100 each month to help pay for the l

aptop. After 5 months, the total amount saved is 75% of the total cost of the laptop. How much more does the student need to save to buy the laptop rounded to the nearest cent.

1 answer:

Andre45 [30]2 years ago

4 0

Answer:

The amount the student needs to save is approximately $269.167

Step-by-step explanation:

The amount the student already has saved for his laptop = $307.50

The amount the student puts aside each month to pay for the laptop = $100

The total amount saved after 5 months = 75% of the total cost of the laptop

Therefore, we have;

The total amount the student has after 5 months = $307.5 + 5 × $100 = $807.5

The total amount saved after 5 months = 75% of the total cost of the laptop

Therefore;

$807.5 = 75% of the total cost of the laptop

The total cost of the laptop = $807.5/75% = $807.5/75 × 100 = $\$ \ 1076\frac{2}{3}$

The total cost of the laptop ≈ $1076.67

Therefore;

The amount the student needs to save = $1076.67 - $807.5 = $\$ \ 269\frac{1}{6}$

The amount the student needs to save ≈ $269.167.

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The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable

topjm [15]

Answer:

(A) 0.15625

(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So, X ~ U(32 , 64)

The PDF of uniform distribution is given by;

f(X) = $\frac{1}{b-a}$ , a < X < b where a = 32 and b = 64

The CDF of Uniform distribution is P(X <= x) = $\frac{x-a}{b-a}$

(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)

P(X > 59) = 1 - P(X <= 59) = 1 - $\frac{x-a}{b-a}$ = 1 - $\frac{59-32}{64-32}$ = $1-\frac{27}{32}$ = 0.15625

(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)

P(X <= 43) = $\frac{43-32}{64-32}$ = $\frac{11}{32}$ = 0.34375

P(X < 37) = $\frac{37-32}{64-32}$ = $\frac{5}{32}$ = 0.15625

P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

(C) Probability that student complete the quiz in exactly 44.74 minutes

= P(X = 44.74)

The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.

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2 years ago

Perpendicular segments are best described as being which of the following

KatRina [158]

It's A. Segments that intersect at the right angle

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2 years ago

Read 2 more answers

Julia is standing 2 feet away from a lamppost. If she casts a shadow of 5 feet and the light makes a 20° angle relative to the g

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Answer:

2.5477 feet

Step-by-step explanation:

Refer the image attached to understand my solution.

BC = height of lamppost.

DE = Julia

AD = shadow of Julia

BD = 2 feet. AD = 5 feet

BA = BD + AD

= 2 + 5 = 7 feet

In ΔABC

tan 20° = $\frac{BC}{BA} = \frac{BC}{7}$

BC = 7 * tan 20°

BC = 2.5477 feet

SO the height of lamppost = 2.5477 feet

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2 years ago

The percent of a country's households with broadband Internet access can be modeled by the function f left parenthesis x right

mojhsa [17]

Answer:

a) The function has a maximum at x = 29,59

b) 99,32 %

Step-by-step explanation:

f(x) = -0,11*x² + 6,51*x + 3

Then:

f´(x) = -0,22*x + 6,51

If f´(x) = 0 then - 0,22*x + 6,51 = 0

0,22*x = 6,51

x = 6,51/0,22 ⇒ x = 29,59

if we get the second derivative

f´´(x) = - 0,22 f´´(x) < 0 then we have a maximun at x = 29,59

a) the function has a maximum

b) 29,59 years after 2000 ( in 2030 )

c) f(x) = - 0,11*x² + 6,51*x + 3

f(29,59) = - (0,11)* (29,59)² + 6,51*29,59 + 3

f(29,59) = 99,32 %

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2 years ago

Suppose that on a certain examination in advanced mathematics, students from univer sity A achieve scores that are normally dist

harkovskaia [24]

Answer:

$P(z>-1.768)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the scores for the univerisity A, and we know that:

$X \sim N (\mu = 625,\sigma =\sqrt{100}= 10)$

Let Y the scores for the univerisity B, and we know that:

$Y \sim N (\mu = 600,\sigma =\sqrt{150}= 12.25)$

We select a sample size of size n=2, and since the distirbution for X is normal then the distribution for the sample mean would be given by:

$\bar X \sim N(\mu=625, \frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{2}}=7.07)$

And for the univeristy B we select a sample of n=3

$\bar Y \sim N(\mu=600, \frac{\sigma}{\sqrt{n}}=\frac{12.25}{\sqrt{3}}=7.07)$

Since both sample means are normally distributed then the difference $Z= \bar X- \bar Y$ is also normal distributed with the following parameters:

$Z= \bar X -\bar Y \sim N(\mu_Z=625-600=25, \sigma_z= \sqrt{100+100}=14.14)$

And we want this probability:

$P(Z>0)$

And we can use the z score given by:

$Z= \frac{z -\mu_z}{\sigma_z}$

And if we replace we got :

$Z= \frac{0-25}{14.14}=-1.768$

And if we find the probability using the normla standard table or excel we got:

$P(z>1.768) =1-P(Z$

3 0

2 years ago