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lakkis [162]
3 years ago
13

A store is having a 15% off sale on all shirts. When Suri checks out, the clerk takes $4.50 off the original price of the shirt.

What was the original cost of the shirt? Use the formula to find the answer.
15% of x = $4.50
Mathematics
1 answer:
Nastasia [14]3 years ago
5 0

Answer: $30

Step-by-step explanation:

Starting with the original formula, you can solve for the cost of the shirt.

First, convert the percentage to a decimal - 15% = .15, and then plug it into the formula:

.15x = $4.50

In order so solve this you need to divide both sides by .15:

X = $30

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Two numbers, a and b, are stored in one byte floating point notation using the least significant (rightmost) 3 bits for the expo
ehidna [41]

a=00110111 = 00110_2 \times 2^{111_2} = 6 \times 2^{-1} = 3


b=11011000 =  11011_2 \times 2^{000_2} = -(0100_2 + 1) =-(4+1)=-5


a+b=-2 = -1 \times 2^{001} = -2 \times 2^{0} = -4 \times 2^{-1} etc.


a+b=-2 = 11111001 = 11110000 = 11100111 = ...


Hard to select the correct answers without seeing the choices. Let's check a couple,


11111001 = -(00000+1) \times 2^{1} = -2 \quad\checkmark


11110000 = -(00001 + 1) \times 2^{0} = -2 \quad\checkmark


11100111 = -(00011 + 1) \times 2^{-1} = -4/2= -2 \quad\checkmark


4 0
3 years ago
I need help with this problem please! I'll add a picture of the problem with the answers.
GaryK [48]
Where is th epicture?
7 0
3 years ago
Read 2 more answers
Twenty-five students from Harry High School were accepted at Magic University. Of those students, 10 were offered athletic schol
Taya2010 [7]

Answer:

Step-by-step explanation:

Part A

For Athletic scholarship,

Mean = (16 + 24 + 20 + 25 + 24 + 23 + 21 + 22 + 20 + 20)/10 = 21.5

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (16 - 21.5)^2 + (24 - 21.5)^2 + (20 - 21.5)^2 + (25 - 21.5)^2 + (24 - 21.5)^2 + (23 - 21.5)^2 + (21 - 21.5)^2 + (22 - 21.5)^2 + (20 - 21.5)^2 + (20 - 21.5)^2 = 64.5

Standard deviation = √64.5/10 = 2.54

For non athletic scholarship,

Mean = (23 + 25 + 26 + 30 + 32 + 26 + 28 + 29 + 26 + 27 + 29 + 27 + 22 + 24 + 25)/15 = 26.6

n = 15

Summation(x - mean)² = (23 - 26.6)^2 + (25 - 26.6)^2 + (26 - 26.6)^2 + (30 - 26.6)^2 + (32 - 26.6)^2 + (26 - 26.6)^2 + (28 - 26.6)^2 + (29 - 26.6)^2 + (26 - 26.6)^2 + (27 - 26.6)^2 + (29 - 26.6)^2 + (27 - 26.6)^2 + (22 - 26.6)^2 + (24 - 26.6)^2 + (25 - 26.6)^2 = 101.6

Standard deviation = √101.6/15 = 2.6

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let 1 be the subscript for scores of athletes and 2 be the subscript for scores of non athletes.

Therefore, the population means would be μ1 and μ2

The random variable is x1 - x2 = difference in the sample mean scores of athletes and non athletes.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 21.5

x2 = 26.6

s1 = 2.54

s2 = 2.6

n1 = 10

n2 = 15

t = (21.5 - 26.6)/√(2.54²/10 + 2.6²/15)

t = - 4.65

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [2.54²/10 + 2.6²/15]²/[(1/10 - 1)(2.54²/10)² + (1/15 - 1)(2.6²/15)²] = 1.2008/0.1039

df = 12

We would determine the probability value from the t test calculator. It becomes

p value = 0.00056

Since alpha, 0.1 > than the p value, 0.00056, then we would reject the null hypothesis.

Therefore, these data provide convincing evidence of a difference in ACT scores between athletes and nonathletes.

Part B

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 90% confidence level, the z score from the normal distribution table is 1.645

x1 - x2 = 21.5 - 26.6 = - 5.1

√(s1²/n1 + s2²/n2) = √(2.54²/10 + 2.6²/15) = 1.05

The confidence interval is - 5.1 ± 1.05

This analysis provides evidence that the mean scores for non athletes is higher than the mean scores for athletes, and that the difference between means in the population is likely to be between 4.05 and 6.15

4 0
3 years ago
Before making invitations, Katrina counted the supplies she bought for her party.
aleksandrvk [35]
The GCF of 36 and 81 is 9

36/9 = 4 water balloons
81/9 = 9 beads

Katrina can invite 9 friends, with each getting 4 water balloons and 9 beads
5 0
2 years ago
Read 2 more answers
You and your friend mix water and cirtic acid. You add 3 cups of citrtic acid for every 16 cups of water Your friend adds 2 cups
Vlad [161]

Answer:

A's mixture is more acidic.

Step-by-step explanation:

Given:

Two friends mix water and citric acid.

One (Let A) add 3 cups of citric acid for every 16 cups of water and another (Let B)adds 2 cups of citric acid for every 12 cups of water.

Now, we have to find that whose mixture is more acidic means:-

Solution:

A add 3 cups of citric acid for every 16 cups of water, the ratio = \frac{3}{16}

B adds 2 cups of citric acid for every 12 cups of water = \frac{2}{12}

Now, to find that who mix more citric acid in the mixture, firstly we will equalize the denominator of both fraction by taking LCM so that can be compare easily.

LCM of 16 and 12:- <u>48</u>

16 - 16, 32, <u>48</u>, 64, 80, 96....

12 - 12, 24, 36, <u>48, </u>60, 72.......

For friend A - \frac{3\times3}{16\times3} =\frac{9}{48}

For friend B - \frac{2\times4}{12\times4} =\frac{8}{48}

<em>We found that in terms of 48 cups of water, friend A mixes 9 cups of acid while friend B mixes 8 cups of acid, means A is adding more acid than B and</em> hence friend A's mixture is more acidic.

4 0
3 years ago
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