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2 years ago

Mary bought 4 gallons of milk for $10.00. Jill bought 2.5 gallons for $6.00.

Mathematics

1 answer:

RoseWind [281]2 years ago

3 0

Step-by-step explanation:

In order to find the better buy, we must find the cheaper unit price or price per gallon. To find the price per gallon, divide the cost by the number of gallons.

Mary

Mary bought 4 gallons for $11.

The price is $11 and the gallons are 4.

Jill

Jill bought 2.5 gallons for $7.25

The price is $7.25 and the gallons are 2.5

Better buy

Mary paid $2.75 per gallon and Jill paid $2.90 per gallon. 2.75 is less than 2.90, so Mary had the better buy.

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What is your sister’s total cost under each of the two plans?2. Suppose your sister doubles her monthly usage to 3,500 minutes a

chubhunter [2.5K]

Answer:

(1) The total cost under Plan A is $92 and the total cost under Plan B is $515.

(2) The total cost under Plan A is $92 and the total cost under Plan B is $1,030.

Step-by-step explanation:

The question is incomplete, so the complete question is below:

$A\ cell \ phone \ company\ offers\ two \ different \ plans.$

$Plan\ A\ costs\ \$92\ per\ month\ for\ unlimited\ talk\ and\ text.\ Plan\ B\ costs \ $0.20\ per\ minute\ plus\ \$0.10\ per\ text\ message\ sent.$ $You\ need \ to\ purchase \ a \ plan \ for\ your \ 14-year-old \ sister$ $.Your\ sister\ currently \ uses\ 1,750 \ minutes\ and \ sends\ 1,650 \ texts\ each \ month.$ $(1) What\ is\ your \ sister's\ total\ cost\ under \ each\ of\ the\ two\ plans?$

$(2) Suppose\ your\ sister\ doubles\ her\ monthly\ usage \ to\ 3,500\ minutes \ and\ sends\ 3,300 \ texts.$ $What \ is \ your \ sister's \ total \ cost \ under \ each \ of \ the \ two \ plans?$

Now, to find (1) total cost for each of the two plans. (2) Total cost under each of the two plans, if sister doubles her monthly usage to 3,500 minutes and sends 3,300 texts.

As, the Plan A is for unlimited talk and text

Cost of the Plan A = $92.

Now, to find the cost under Plan B:

According to question:

Rate of call per minute is $0.20 and per text is $0.10.

Calls she uses currently is 1,750 minutes and text 1,650.

So, to get the cost of Plan B:

$0.20\times 1750+0.10\times 1650$

$=350+165$

$=515.$

Thus, Plan B costs $515.

Now, to get the total cost as, sister doubles her monthly usage to 3,500 minutes and sends 3,300 texts.

As, the Plan A is same in both cases and is for unlimited text and calls.

So, cost of Plan A = $92.

As, the monthly usage is double.

Calls in minutes are 3,500.

Texts are 3,300.

Now, to get the total cost under Plan B:

$0.20\times 3500+0.10\times 3300$

$=700+330$

$=1030.$

Hence, the cost of Plan B = $1,030.

Therefore, (1) The total cost under Plan A is $92 and the total cost under Plan B is $515.

(2) The total cost under Plan A is $92 and the total cost under Plan B is $1,030.

7 0

3 years ago

the measure of two complimentary angles is in the ratio 1:9. what are the degree measures of the two angles

marusya05 [52]

$x+9x=90\\
10x=90\\
x=9\\
9x=9\cdot9=81$

9 and 81.

4 0

3 years ago

Which postulate or theorem can be used to

sveticcg [70]

Answer:

AA Similarity Postulate

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

step 1

Verify the proportion of the corresponding sides

$\frac{PS}{PQ}=\frac{PT}{PR}$

substitute

$\frac{45}{20}=\frac{36}{16}$

$2.25=2.25$ ----> is true

Corresponding sides are proportional

Triangle PQR is similar to Triangle PST

That means

Corresponding angles must be congruent

side QR is parallel side ST

and

$m\angle PQR=m\angle PST$ ----> by corresponding angles

$m\angle PRQ=m\angle PTS$ --> by corresponding angles

PQR is similar to PST by AA Similarity Postulate

4 0

3 years ago

Can anybody help me please

Sliva [168]

Hello, your answer would be (-5, 5)

3 0

2 years ago

A population of 200 animals is decreasing by a rate of .96 per year . At this rate, in how many years will the population be les

mojhsa [17]

At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ

Now, we need to know when the population will be less than 170.

So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n

n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176

7 0

3 years ago