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antiseptic1488 [7]
2 years ago
9

PLSSSS HELP ASAP I AM TIMED. A community garden contains five types of plants. Two of the types are flowers: roses and daisies.T

he other three types are vegetable plants: cucumbers, bell peppers, and green beans.
Plants in the Community Garden
Plant
Number
Roses
2
Daisies
4
Cucumbers
6
Bell Peppers
5
Green Beans
7

Which statetement is true?
Green bean plants are One-third of the total plants.
The ratio of cucumber plants to flowers is 1:1.
Bell pepper plants are One-third of the vegetable plants.
The ratio of daisy plants to rose plants is 1:2.
Mathematics
2 answers:
Ann [662]2 years ago
6 0
I’m sure it’s the second answer because there are 6 cucumber plants and 6 flowers. :)
Mariulka [41]2 years ago
3 0

Answer:

B

Step-by-step explanation:

I did the test (; on egunity

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Price-club sells a 12-ounce box of crackers for $2.59 and Shop Mart sells 1-1/2 pound box of crackers for $4.68. What is the uni
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Unit price would be the price per ounce.

Divide the total cost by the number of ounces:

Price Club : 2.59 / 12 = $0.216 per ounce

Shop Mart: 1-1/2 pounds = 1.5 x 16 = 24 ounces.

4.68 / 24 = $0.195 per ounce.

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the amount of money spent weekly on cleaning maintenance and repairs at a large restaurant was observed over a long period of ti
soldi70 [24.7K]

Answer:


Step-Question: The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was ...

The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean ? = $615 and standard deviation ? = $38.

(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Round your answer to four decimal places.)

(b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only0.08? (Round your answer to the nearest dollar.)

$by-step explanation:


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3 years ago
If a pair of 6-sided dice is tossed, what is the probability that both dice will show a number that is a multiple of 3 (i.E. 3 o
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There are 4 possibilities out of the 36 ways the dice can land:

... (3, 3), (3, 6), (6, 3), (6, 6)

Hence, the probability of an outcome in this set is 4/36 = 1/9.

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Please help and explain how you got the answer. Im giving out brainliest. || Q4
rosijanka [135]

Answer: 32

Step-by-step explanation:

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2 years ago
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A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. ​(a) Describe the
Ivanshal [37]

Answer:

a) \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

b) z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

c) z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

d) z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

\mu = 83, \sigma = 27

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

Part b

We want this probability:

P(\bar X>89)

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 89 we got:

z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

Part c

P(\bar X

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 75.65 we got:

z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

Part d

We want this probability:

P(79.4 < \bar X < 89.3)

We find the z scores:

z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

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3 years ago
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