Solve for X in the figure below. Show your work. 22* Degrees 3x+11

n winter, the price of apples suddenly went up by \$0.75$0.75dollar sign, 0, point, 75 per pound. Sam bought 333 pounds of apple

RSB [31]

$1.21 is the original price per pound...i too am having trouble with figuring out how to setup the equation!!

5 0

2 years ago

A rice merchant purchased 50 bags of rice at rupees 300 per back he spent 500 rupees towards transportation due to lack of deman

Alex777 [14]

Answer:

10500 rupees

Step-by-step explanation:

So 50 bags 300 per pack so

300*50= 15000

15000 is what he spent on the 50 bags of rice

So he spent 500 rupees on transportation

15000 + 500 = 15500 total spent which is not needed but extra information

30% loss is 30% of 15000 so

30% of 15000 = 4500

15000 - 4500 = 10500 rupees

8 0

2 years ago

HELP PLEASE HELP!!!!! 20 POINTS!!!

ivann1987 [24]

The answer should be C. By combining like terms you can find the answer. 1.75a+2.75a = 4.5a
2.25b + 1.75b = 4b
2.25c + 1.25c = 3.5c

3 0

2 years ago

Read 2 more answers

Every day, Jorge buys a lottery ticket. Each ticket has a probability of of winning a prize. After six days, what is the probabi

Romashka [77]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Every day, Jorge buys a lottery ticket. Each ticket has a 0.16 probability of winning a prize. After six days, what is the probability that Jorge has won at least one prize? Round your answer to four decimal places.

Answer:

The probability that Jorge has won at least one prize after six days is

P(at least 1 win) = 0.6487

Step-by-step explanation:

Every day, Jorge buys a lottery ticket which has a 0.16 chance of winning a prize.

We want to find out the probability that Jorge has won at least one prize after six days.

P(at least 1 win) = 1 - P(not winning for 6 days)

We know that the probability of winning is 0.16 then the probability of not winning is

P(not winning) = 1 - 0.16 = 0.84

For 6 days,

P(not winning for 6 days) = 0.84×0.84×0.84×0.84×0.84×0.84

P(not winning for 6 days) = 0.84⁶

P(not winning for 6 days) = 0.3513

Finally,

P(at least 1 win) = 1 - P(not winning for 6 days)

P(at least 1 win) = 1 - 0.3513

P(at least 1 win) = 0.6487

6 0

2 years ago

Louden County Wildlife Conservancy counts butterflies each year. Data over the last three years regarding four types of butterfl

MrRa [10]

Incomple question. However, here's the remaining part of the question:

14

2009

Meadow Fritillary= 5

Variegated Fritillary= 7

Zebra Swallowtail= 33

Eastern-Tailed Blue= 242

Louden County Butterfly Count

2010

Meadow Fritillary 34

Variegated Fritillary 95

Zebra Swallowtail 21

Eastern-Tailed Blue 168

2011

Meadow Fritillary

Variegated Fritillary

Zebra Swallowtail

Eastern-Tailed Blue

10

170

<u>Options</u>:

A) All butterfly populations are steadily decreasing.

B)All butterfly populations were larger than usual in 2010.

C)The Eastern-Tailed Blue butterfly is more common than the others.

D)The Meadow Fritillary is equally common as the Variegated Fritillary

Answer:

<u>C</u>

Step-by-step explanation:

Looking through the above count data by Louden County Wildlife Conservancy from 2009 to 2011 we notice the Eastern- Trailed Blue butterfly has a higher count, which implies that the Eastern-Tailed Blue butterfly is more common than the other butterflies.

Therefore, we could infer from the samples, that the Eastern-Tailed Blue butterfly is more common than others from the records of the past 3 three years.

5 0

2 years ago