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Greeley [361]
3 years ago
12

Shaylee counts 4 boxes of wheat cereal in the cupboard. Each box contains 52 ounces of cereal. She also counts 5 boxes of rice c

ereal. Each box contains 45 ounces of cereal. How many ounces of cereal are in Shaylee’s cupboard?
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
5 0
She has 433 onces of cereal all together
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Suppose 12% of the students chose to study French their freshman year, and that meant that there were 21 such students. How many
AleksAgata [21]

Answer:

154 students

Step-by-step explanation:

First get the total number of students .

This can be gotten by

12% of A = 21

Where A represents the total number of students.

12% represents the % of A that chose to study French and 21 is the number of students that studied French .

Therefore,

12% /100% x A = 21

0.12 x A = 21

Divide both sides by 0.12

0.12/0.12 x A = 21/0.12

A = 175

The total number of students is 175.

If 21 chose to study French their freshman year ,number of students that chose not to will be total number of students minus number of those who chose to study French.

That’s

175 - 21

= 154

154 students chose not to study French their freshman year

4 0
3 years ago
A group of veterinarians at a major veterinary hospital was interested in investigating a possible link between enteroliths, sto
klemol [59]
C. all horses that are diagnosed with enteroliths
3 0
3 years ago
What is -1 = 5/8b + 3/8
andrew-mc [135]

Answer:

exact form - b = -11/5

decimal form - b = -2.2

Mixed number form - b = -2 1/5

4 0
3 years ago
The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is
marysya [2.9K]

Answer:

(a) 0.14%

(b) 2.28%

(c) 48%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let <em>X</em> be a random variable representing the prices paid for a particular model of HD television.

It is provided that <em>X</em> follows a normal distribution with mean, <em>μ</em> = $1600 and standard deviation, <em>σ</em> = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})

                   =P(Z>3)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

P(X

                   =P(Z

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

P(1400

                              =P(-2

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

P(1500

                              =P(-1

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

P(1600

                              =P(0

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

P(1600

                              =P(0

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.

8 0
3 years ago
PLEASE HELP
Alexeev081 [22]
Its may be A or C
i am not sure just try to help
4 0
3 years ago
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