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2 years ago

3

Mrs Frasier is making costumes for the school play. Each costume requires 0.75 yard of fabric. she bought 6 yards of fabric. how

many costumes can sh make?

2 answers:

Sonbull [250]2 years ago

5 0

12 I think is the answer but if its not sorry

nikdorinn [45]2 years ago

4 0

You get 6 and divide by .75 and your answer is 8 costumes

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Complete the area model representing the polynomial x2 − 11x + 28. What is the factored form of the polynomial? Use the model to

qwelly [4]

Answer:

$\boxed{\sf \ \ \ (x-4)(x-7) \ \ \ }$

Step-by-step explanation:

Hello,

from the expression we can say that the sum of the two zeroes is 11 and the product is 28

4 + 7 = 11

4 * 7 = 28

so we can write

$x^2-11x+28=(x-4)(x-7)$

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2 years ago

Estimates show that there are 1.4x10^8 pet fish and 9.4x10^6 pet reptiles in the United States. How many pet fish and reptile pe

azamat

Answer:

There are total number of pet fish and reptile pets are $1.49\times 10^8$ .

Step-by-step explanation:

Number of pet fish are $1.4\times 10^8$ and the number of pet reptiles are $9.4\times 10^6$ .

It is required to find the total number of pet fish and reptile pets in the United States. It can be done by simply adding the above numbers such that total numbers are given by :

$T=1.4\times10^{8}+9.4\times10^{6}\\\\T=149400000\\\\T=1.49\times 10^8$

So, there are total number of pet fish and reptile pets are $1.49\times 10^8$ .

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2 years ago

Bill's favorite breakfast cereal is available in four different sizes. Which is the best buy?

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2 years ago

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Suppose babies born in a large hospital have a mean weight of 3316 grams, and a standard deviation of 324 grams. If 83 babies ar

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Answer: 0.129

Step-by-step explanation:

Let $\overline{X}$ denotes a random variable that represents the mean weight of babies born.

Population mean : $\mu= \text{3316 grams,}$

Standard deviation: $\text{324 grams}$

Sample size = 83

Now, the probability that the mean weight of the sample babies would differ from the population mean by greater than 54 grams will be :

$P(|\mu-\overline{X}|>54)=1-P(\dfrac{-54}{\dfrac{324}{\sqrt{83}}}$

hence, the required probability = 0.129

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2 years ago

mars, inc. claims that 20% of its m&m plain candies are orange. a sample of 100 such candies is randomly selected. find t

elena-14-01-66 [18.8K]

The probability p of an orangecandy is 0.2. The sample size = 100.
The mean is given by:
$n\times p=100\times0.2=20$
The standard deviation is given by:
$\sqrt{npq} = \sqrt{100\times0.2\times(1-0.2)} =4$
The answers are: Mean = 20. Standard deviation = 4.

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2 years ago

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