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2 years ago

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james bought a cheeseburger, fries, and a drink for dinner. the cheeseburger was three times the price of the fries and the drin

k and the fries were the same price. if the entire meal was 12.50, what was the price of each item

1 answer:

nordsb [41]2 years ago

7 0

The answer is $4.16 for each item

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Sven is trying to find the maximum amount of time he can spend practicing the five scales of piano music he is supposed to be wo

Katena32 [7]

Answer:

He should spend 3 minutes or less on each scale

Sven made a mistake in the symbol of inequality, placing lesser or equal instead of greater or equal

Step-by-step explanation:

Let

t ------> is the number of minutes he spends on each scale

Remember that the phrase "at least" is equal to "greater than or equal"

so

The inequality that represent this scenario is

$70-5t \geq 55$

solve for t

$-5t \geq 55-70$

$-5t \geq -15$

Multiply by -1 both sides

$5t \leq 15$

Divide by 5 both sides

$t \leq 3$

Sven is incorrect

He should spend 3 minutes or less on each scale

Sven made a mistake in the symbol of inequality, placing lesser or equal instead of greater or equal

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2 years ago

A recipe for 11 pizza crust calls for 3838 tablespoon of salt. How much salt is needed to make 33 pizza crusts?

max2010maxim [7]

Its 11,514 tablespoons of salt

8 0

2 years ago

a photo collage consists of seven equal sized square photo arranged in a row to form a rectangle if the total area of the colleg

Keith_Richards [23]

Answer:

Step-by-step explanation:

I have attached a written explanation to help :).

Let's say each side of the square = $x$

If we put 7 squares in a row to make the rectangle,

length of the rectangle = $7x$

Area of a rectangle = length multiplied by width

Area of rectangle = $7x$ multiplied by $x$ = $7x^2$

We are also told that the total area = 567

So $7x^2 = 567$

Now solve for $x$ .

Divide both sides by 7.

$\frac{7x^2}{7} = \frac{567}{7}$

$x^2 = 81$

Then square root both sides to find $x$

$x = \sqrt{81}$ = 9

Finally, the question asks to find the length of the collage which we said was $7x$ and we know $x = 9$

Length of collage = $7x = 7$ multiplied by $9$ = ?

7 0

2 years ago

1. The following are the number of hours that 10 police officers have spent being trained in how to handle encounters with peopl

dusya [7]

Answer:

$Range = 16$

$Inter\ Quartile\ Range = 6.75$

$Variance = 20.44$

$Standard\ Deviation = 4.52$

Step-by-step explanation:

Given

4, 17, 12, 9, 6, 10, 1, 5, 9, 3

Calculating the range;

$Range = Highest - Lowest$

From the given data;

Highest = 17 and Lowest = 1

Hence;

$Range = 17 - 1$

$Range = 16$

Calculating the Inter-quartile Range

Inter quartile range (IQR) is calculates as thus

$IQR = Q_3 - Q_1$

Where

Q3 = Upper Quartile and Q1 = Lower Quartile

<em />

<em>Start by arranging the data in ascending order</em>

1, 3, 4, 5, 6, 9, 9, 10, 12, 17

N = Number of data; N = 10

---------------------------------------------------------------------------------

Calculating Q3

$Q_3 = \frac{3}{4}(N+1) th\ item$

<em>Substitute 10 for N</em>

$Q_3 = \frac{3}{4}(10+1) th\ item$

$Q_3 = \frac{3}{4}(11) th\ item$

$Q_3 = \frac{33}{4} th\ item$

$Q_3 = 8.25 th\ item$

Express 8.25 as 8 + 0.25

$Q_3 = (8 + 0.25) th\ item$

$Q_3 = 8th\ item + 0.25 th\ item$

Express 0.25 as fraction

$Q_3 = 8th\ item +\frac{1}{4} th\ item$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

From the arranged data;

$8th\ item = 10$ and $9th\ item = 12$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

$Q_3 = 10 +\frac{1}{4} (12 - 10)$

$Q_3 = 10 +\frac{1}{4} (2)$

$Q_3 = 10 +0.5$

$Q_3 = 10.5$

Calculating Q1

$Q_1 = \frac{1}{4}(N+1) th\ item$

<em>Substitute 10 for N</em>

$Q_1 = \frac{1}{4}(10+1) th\ item$

$Q_1 = \frac{1}{4}(11) th\ item$

$Q_1 = \frac{11}{4} th\ item$

$Q_1 = 2.75 th\ item$

Express 2.75 as 2 + 0.75

$Q_1 = (2 + 0.75) th\ item$

$Q_1 = 2nd\ item + 0.75 th\ item$

Express 0.75 as fraction

$Q_1 = 2nd\ item +\frac{3}{4} th\ item$

$Q_1 = 2nd\ item +\frac{3}{4} (3rd\ item - 2nd\ item)$

From the arranged data;

$2nd\ item = 3$ and $3rd\ item = 4$

$Q_1 = 3 +\frac{3}{4} (4 - 3)$

$Q_1 = 3 +\frac{3}{4} (1)$

$Q_1 = 3 +0.75$

$Q_1 = 3 .75$

---------------------------------------------------------------------------------

Recall that

$IQR = Q_3 - Q_1$

$IQR = 10.5 - 3.75$

$IQR = 6.75$

Calculating Variance

Start by calculating the mean

$Mean = \frac{1+3+4+5+6+9+9+10+12+17}{10}$

$Mean = \frac{76}{10}$

$Mean = 7.6$

Subtract the mean from each data, then square the result

$(1 - 7.6)^2 = (-6.6)^2 = 43.56$

$(3 - 7.6)^2 = (-4.6)^2 = 21.16$

$(4 - 7.6)^2 = (-3.6)^2 = 12.96$

$(5 - 7.6)^2 = (-2.6)^2 = 6.76$

$(6 - 7.6)^2 = (-1.6)^2 = 2.56$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(10 - 7.6)^2 = (2.4)^2 = 5.76$

$(12 - 7.6)^2 = (4.4)^2 = 19.36$

$(17 - 7.6)^2 = (9.4)^2 = 88.36$

Sum the result

$43.56 + 21.16 + 12.96 + 6.76 + 2.56 + 1.96 + 1.96 + 5.76 + 19.36 + 88.36 = 204.4$

Divide by number of observation;

$Variance = \frac{204.4}{10}$

$Variance = 20.44$

Calculating Standard Deviation (SD)

$SD = \sqrt{Variance}$

$SD = \sqrt{20.44}$

$SD = 4.52$ <em>(Approximated)</em>

4 0

2 years ago

Determine whether the sets have a subset relationship. R = {right triangles} O = {obtuse triangles} a) R and O are disjoint. b)

DerKrebs [107]

Answer:

f

Step-by-step explanation:

A right angled triangle has one angle equal to 90°

an obtuse angle triangle has one angle greater tan 90°

a) An o and right-angled triangle btuse angled triangle can have one angle equal.

b) R and O are not equivalent by definition

c) subset of R and Oare not all triangles as R and O are categories of all triangles. R and O are subset of all triangles

d) Acute triangles have all acute angles. So subset of R and O can't be all acute traingles

e) All triangles with two acute angles are may or may not have third angle as obtuse angle or 90° angle

f) none of the above

5 0

2 years ago