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lidiya [134]
2 years ago
10

Lisa has $7.80 to spend on some tomatoes and a loaf of bread. Tomatoes cost $1.20 per pound, and a loaf of bread costs $1.80.The

inequality 1.20x + 1.80 ≤ 7.80 models this situation, where x is the number of pounds of tomatoes.Solve the inequality. How many pounds of tomatoes can Lisa buy
Mathematics
1 answer:
kenny6666 [7]2 years ago
4 0

Answer:

At maximum, she can buy 5 tomatoes.

Step-by-step explanation:

If you solve the inequality, as so:

1.2x+1.8 <=7.8

1.2x <= 6

x <= 5

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Ok these are the last ones
Nutka1998 [239]

There are 12 months in a year

4 months that have 30 days so ratio of the months with 30 days to the  months with other than 30 days = 4: 12 = 1:3

Answer:

1 to 3

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

24 sedans and 18 sports

Ratio of sedans to sport = 24 : 18 = 4:3

Answer:

4 to 3

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

30 inches to 2 feet  

2 feet = 24 inches

so 30 inches to 24 inches = 30:24

simplify

= 5 : 4

Answer:

5 to 4

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

36 inches to 4 feet = 36 inches to 48 inches = 36:48

simplify

= 3 : 4  

Answer:

3 to 4

6 0
3 years ago
Read 2 more answers
Suppose that the cost of a rental car is $65 for a week plus 0.15 per mile driven.
IceJOKER [234]

You have the following equation for the cost of renting a car for x number of miles driven:

y = 0.15x + 65

If you travel 70 miles, then, x = 70 and you obtain for the cost y:

y = 0.15(70) + 65

y = 10.5 + 65

y = 75.5

Hence, the cost of renting a car to drive 70 miles is $75.5

3 0
1 year ago
Don't be rude. Answer the questoin .<br>​
MAXImum [283]

Answer:

Step-by-step explanation:

3. sin^2A + cos^2A = 1  (formula )

1 + tan^2 theta = sec^2 theta  (derived from the formula sec^2 theta - tan^2 theta )

b.2x = 12 (by comparing the corresponding element)

x = 12/2

x = 6

3y = 12 (by comparing the corresponding element)

y = 12/3

y = 4

mean = sum of data / no of data

4+7+5+9+11+6 / 6

42 / 6

7

mean = 7

to find median arrange the data from ascending to descending order (from smallest to biggest)

data = 3,4,6,8,10,12

median = (N +1)/2   (N means no of data )

=6 + 1/2

=7/2

=3.5

=3 th term + 4 th term /2

=6 +8 / 2

=14/2

=7 median = 7

to find value of theta

90° + 50° + theta = 180° (sum of interior angles of a triangle)

140° + theta = 180°

theta = 180° - 140°

theta = 40°

therefore theta = 40°

Hope this helps u !!

4 0
3 years ago
The box plots compare the number of calories in each snack pack of crackers and cookies.
Nikitich [7]

Answer:

4th statement is true.

Step-by-step explanation:

We have been two box plots, which represents the number of calories in each snack pack of crackers and cookies. We are asked to find the correct statement about our given box plots.

1. More packets of crackers have 80 calories than any other number of calories.

We can see that median of box plot representing calories of cookies is 80. This means that half of the packets of crackers have less than 80 calories and half of the packets have more than 80 calories, therefore, 1st statement is false.

2. The value 70 is an outlier for the number of calories in the cookie pack.

Since an outlier is 1.5 times the interquartile range.

IQR=Q_3-Q_1

\text{IQR of cookie packs}=105-90

\text{IQR of cookie packs}=15

\text{Lower outlier}=Q_1-(1.5*IQR)

\text{Lower outlier}=90-(1.5*15)

\text{Lower outlier}=90-22.5

\text{Lower outlier}=67.5

Since any number less than 67.5 will be an outlier and 70 is grater than 67.5, therefore, 70 is not an outlier in number of calories in cookie packs and 2nd statement is false.

3. The upper quartile of the cookie data is equivalent to the maximum in the cracker data.

We can see that upper quartile of cookie data is 105 and the maximum in cracker data is 100. Since 105 is greater than 100, therefore, 3rd statement is false.

4. The number of calories in each pack of cookies has a greater variation than the number of calories in each pack of crackers.

Since range and IQR are good measures of variation of box-plots, so we will find the range and IQR of our both box-plots.

We have already seen that IQR of cookie packs is 15.

\text{IQR of cracker packs}=85-75

\text{IQR of cracker packs}=10

\text{Range}=\text{Maximum value - Minimum value}

\text{Range of calories in cracker packs}=100-70

\text{Range of calories in cracker packs}=30

\text{Range of calories in cookie packs}=115-70

\text{Range of calories in cookie packs}=45  

We can see that the range of calories in cookie packs (45) is greater than range of calories in cracker packs (30) and IQR of calories in cookie packs (15) is greater than IQR of calories in cracker packs (10), therefore, 4th statement is true.

3 0
4 years ago
Read 2 more answers
What is the nth term for 7,11,15,19
Murrr4er [49]

Answer:

a_{n} = 4n + 3

Step-by-step explanation:

Note there is a common difference d between consecutive terms, that is

d = 11 - 7 = 15 - 11 = 19 - 15 = 4

This indicates the sequence is arithmetic with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 7 and d = 4, thus

a_{n} = 7 + 4(n - 1) = 7 + 4n - 4 = 4n + 3

5 0
3 years ago
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