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4 years ago

15

Gail is making costumes for a school play. Each rabbit costume needs one and one half yards of white fur fabric, a yard of blue

striped fabric, and a quarter of a yard of pink felt for the ears. G?ail needs to make eight rabbit costumes.??How man yards will she need, for the white fur fabric, blue striped fabric, and pink felt?

2 answers:

olga nikolaevna [1]4 years ago

7 0

Answer:

Gail will need 12 yards of white fur, 8 yards of blue striped and 2 yards of the pink felt.

Step-by-step explanation:

We are given that,

The amount of yards for making one costume of different fabrics are,

White fur fabric = $1\frac{1}{2}$ = $\frac{3}{2}$ yards

Blue striped fabric = 1 yard

Pink felt for ears = $\frac{1}{4}$ yards.

So, we have,

The amount required to make 8 costumes are,

White fur fabric = $\frac{3}{2}\times 8$ = 3×4 = 12 yards

Blue striped fabric = 1×8 = 8 yards

Pink felt for ears = $\frac{1}{4}\times 8$ = 2 yards.

Thus, Gail will need 12 yards of white fur, 8 yards of blue striped and 2 yards of the pink felt.

Arte-miy333 [17]4 years ago

5 0

Answer: There are 12 yards of white fur fabric , 8 yards of blue striped fabric and 2 yards of pink she needed for 8 rabbit costumes.

Step-by-step explanation:

Since we have given that

Number of yards of white fur fabric each rabbit needs is given by

$1\frac{1}{2}=\frac{3}{2}\ yards$

Number of yard of blue striped fabric each rabbit needs = 1

Number of yard of pick felt for the ears is given by

$\frac{1}{4}$

Total number of rabbit costume they needed = 8

Number of yards of white fur fabric for total number of rabbit costume she needs is given by

$8\times \frac{3}{2}=4\times 3=12\ yards$

Number of yards of blue striped fabric for total number of rabbit costume she needed is given by

$8\times 1\\\\=8\ yards$

Number of yards of pink felt for the ears for total number of rabbit costume she needed is given by

$\frac{1}{4}\times 8\\\\=2\ yards$

Hence, there are 12 yards of white fur fabric , 8 yards of blue striped fabric and 2 yards of pink she needed for 8 rabbit costumes.

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The slope of RS, RT, and ST are 1/2, -1 and -4 respectively

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Given the folllowing coordinates R (2,3), S (4,4), and T (5,0) os triangle RST

The formula for calculating the distance is expressed as:

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For length RS:

D = √(4-3)²+(4-2)²

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For length RS:

RT = √(0-3)²+(5-2)²

RT = √9+9
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For length ST:

D = √(0-4)²+(5-4)²

ST = √16+1
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Since all the sides are different, the triangle is a scalene triangle.

For the slope of the sides

For the side RS;

Slope = 4-3/4-2

Slope of RS = 1/2

For the side RT:

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For the side ST

Slope of ST = 0-4/5-4
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Hence the slope of RS, RT, and ST are 1/2, -1 and -4 respectively

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Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard

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Answer:

a) 36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b) 66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c)

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Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 72, \sigma = 6.5$

a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 66 beats per minute and 78 beats per minute.

The probability is?

This is the pvalue of Z when X = 78 subtracted by the pvalue of Z when X = 66. So

X = 78

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{78 - 72}{12.5}$

$Z = 0.48$

$Z = 0.48$ has a pvalue of 0.6844

X = 66

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{66 - 72}{12.5}$

$Z = -0.48$

$Z = -0.48$ has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b. If 4 adult females are randomly selected, find the probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

The probability is?

Now we have $n = 4, s = \frac{12.5}{\sqrt{4}} = 6.25$

So

X = 78

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{78 - 72}{6.25}$

$Z = 0.96$

$Z = 0.96$ has a pvalue of 0.8315

X = 66

$Z = \frac{X - \mu}{s}$

$Z = \frac{66 - 72}{6.25}$

$Z = -0.96$

$Z = -0.96$ has a pvalue of 0.1685

0.8315 - 0.1685 = 0.6630

66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The condition for the sample size exceeding 30 is when the population is skewed. If it is normally distributed, the size is not a condition.

So the correct answer is:

C.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

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