MATHEMATICS COLLEGE

Someone please help??

Answers

Answer 1

Answer:

9514 1404 393

Answer:

Isabel

Step-by-step explanation:

Gavin saves 28% (given).

Harry saves 1 -3/4 = 1/4 = (1/4)·100% = 25%.

Isabel saves 3/(3+7)·100% = 30%.

Isabel saves the largest fraction of her salary each month.

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Add the numbers in the series 3+11+19+27+.....+395+403.

Answers

Answer:

10353

Step-by-step explanation:

The given series is in arithmetic progression since the common difference is same which is 8.

To find the sum of series we can simply apply the formula'

S= n/2( first term + last term)

S is the sum and n is the number of terms

we also need to find the number of terms n

n = (last term- first term)/2 + 1

$n= (403-3)/(8) + 1$

n= 51

$s= (51)/(2)(3+403)$

s= 10353

The only swimming pool at the El Cheapo Motel is outdoors. It is 5.0 m wide and 12.0 m long. If the weekly evaporation is 2.35 in., how many gallons of water must be added to the pool if it does not rain?

Answers

Answer:

946.10 gallons per week

Step-by-step explanation:

1 cm = 0.393701 inch

Width = 5.0m = 196.85 inch

Length = 12.0 m = 472.44 inch

The volume evaporated weekly is given by:

$V = L*W*2.35 = 196.85*472.33*2.35\nV=218,550.12\ in^3$

Converting to gallons:

$1\ gal = 231\ in^3\nV=(218,550.12)/(231)\nV= 946.10\ gal$

946.10 gallons of water must be added to the pool each week.

Final answer:

To determine the volume of water evaporated from the pool at the El Cheapo Motel, we converted all measurements to a common unit and calculated the volume of water evaporated. The motel has to add approximately 946.13 gallons of water weekly, considering there is no rain.

Explanation:

To answer this question, we first need to convert the measurements to a common unit. Given that the pool is 5.0 m wide and 12.0 m long (a total area of 60.0 m2) and the weekly evaporation is 2.35 inches, we first convert the inches to meters. Since 1 inch is equal to 0.0254 meters, 2.35 inches equals 0.05969 meters.

Then, we calculate the volume of water evaporated in a week, which is calculated by multiplying the surface area of the pool by the depth of the water evaporated. Hence, it's 60.0 m2 * 0.05969 m = 3.58 m3. As 1 m3 is approximately 264.17 gallons, 3.58 m3 equals 946.1296 gallons approximately.

In conclusion, the El Cheapo Motel needs to add around 946.13 gallons of water to their pool on a weekly basis, if there is no rain.

Learn more about Volume Calculation here:

brainly.com/question/32822827

#SPJ3

The quadratic function f(x) has a vertex at (9, 8) and opens upward. If g(x) = 4(x − 8)2 + 9, which statement is true?A.
The maximum value of f(x) is greater than the maximum value of g(x).
B.
The maximum value of g(x) is greater than the maximum value of f(x).
C.
The minimum value of f(x) is greater than the minimum value of g(x).
D.
The minimum value of g(x) is greater than the minimum value of f(x).

Answers

ANSWER

D.
The minimum value of g(x) is greater than the minimum value of f(x).

EXPLANATION

It was given that;

$f(x)$
has a vertex at

$(9,8)$

and opens upwards.

This means that f(x) is a minimum graph and hence have a minimum value of 8.

Also ,

$g(x) = 4 {(x - 8)}^(2) + 9$

When we compare this function to

$y = a {(x - h)}^(2) + k$

We can see that,a=4, h=8 and y=9.

The vertex is

$(8,9)$

Since a>0, the graph opens upwards.

The graph has a minimum point which is (8,9) and hence the minimum value is 9.

We can see that, the minimum value of g(x) is greater than the minimum value of f(x).

Therefore the correct answer is D.

In determining automobile-mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following:__________. a. P(X<30)
b. P(28c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.

Answers

The upper 5% of cars have a mileage rating of 35.805 mpg

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given; mean of 33 mpg and a standard deviation of 1.7

a) For < 30:

z = (30 - 33)/1.7 = -1.76

P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392

b) For < 28:

z = (28 - 33)/1.7 = -2.94

P(x < 28) = P(z < -2.94) = 0.0016

c) For > 35:

z = (35 - 33)/1.7 = 1.18

P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119

d) For > 31:

z = (31 - 33)/1.7 = -1.18

P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810

e) The upper 5% of cars achieve have a z score of 1.65, hence:

1.65 = (x - 33)/1.7

x = 35.805 mpg

The upper 5% of cars have a mileage rating of 35.805 mpg

Find out more on z score at: brainly.com/question/25638875

Answer:

a) P(X < 30) = 0.0392.

b) P(28 < X < 32) = 0.2760

c) P(X > 35) = 0.1190

d) P(X > 31) = 0.8810

e) At least 35.7965 mpg

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 33, \sigma = 1.7$

a. P(X<30)

This is the pvalue of Z when X = 30. So

$Z = (X - \mu)/(\sigma)$

$Z = (30 - 33)/(1.7)$

$Z = -1.76$

$Z = -1.76$ has a pvalue of 0.0392.

Then

P(X < 30) = 0.0392.

b) P(28 < X < 32)

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So

X = 32

$Z = (X - \mu)/(\sigma)$

$Z = (32 - 33)/(1.7)$

$Z = -0.59$

$Z = -0.59$ has a pvalue of 0.2776.

X = 28

$Z = (X - \mu)/(\sigma)$

$Z = (28 - 33)/(1.7)$

$Z = -2.94$

$Z = -2.94$ has a pvalue of 0.0016.

0.2776 - 0.0016 = 0.2760.

P(28 < X < 32) = 0.2760

c) P(X>35)

This is 1 subtracted by the pvalue of Z when X = 35. So

$Z = (X - \mu)/(\sigma)$

$Z = (35 - 33)/(1.7)$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

1 - 0.8810 = 0.1190

P(X > 35) = 0.1190

d. P(X>31)

This is 1 subtracted by the pvalue of Z when X = 31. So

$Z = (X - \mu)/(\sigma)$

$Z = (31 - 33)/(1.7)$

$Z = -1.18$

$Z = -1.18$ has a pvalue of 0.1190.

1 - 0.1190 = 0.8810

P(X > 31) = 0.8810

e. the mileage rating that the upper 5% of cars achieve.

At least the 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then

$Z = (X - \mu)/(\sigma)$

$1.645 = (X - 33)/(1.7)$

$X - 33 = 1.645*1.7$

$X = 35.7965$

At least 35.7965 mpg

Solve for the area of ΔABC to the nearest whole number. A) 13 cm2 B) 26 cm2 C) 53 cm2 D) 106 cm2

Answers

Answer:

The answer is 53 cm2 or c.

[Calc1] Help with 2 questions?

(B) What is the least squares regression line

(C) According to the model in (b), for every increase of $1000 in income, the ulcer rate (per 100 population) will go down by _________ points.

Answers

Answer:

B) y = -9.98×10⁻⁵ x + 13.95

C) 0.1

Step-by-step explanation:

Using Excel, the least squares regression line, rounded to two decimal places, is y = -9.98×10⁻⁵ x + 13.95.

The slope is -9.98×10⁻⁵, so if x increases by 1000, then y changes by -9.98×10⁻² = -0.0998. Rounded to one decimal place, the ulcer rate per 100 population will go down by 0.1 points.

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