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Leya [2.2K]
2 years ago
8

Jack's backyard has a grassy lawn, a deck, and a pool. the pool takes up 1/5 for the yard, the deck takes up 1/4 of the yard the

rest of the yard is left for grass. what fraction of the backyard does the grassy lawn take up?
(sorry my sister asked me for help and i have no clue what it's talking about)
Mathematics
1 answer:
Tatiana [17]2 years ago
5 0

Answer:

11/20

Step-by-step explanation:

The backyard is 1 entire area.

The pool is 1/5 of the area.

The deck is 1/4 of the area.

The grass is the remaining area.

The area of the grass is the entire area of the backyard minus the areas of the pool and the deck.

First, we add the areas of the pool and the deck.

1/4 + 1/5 = 5/20 + 4/20 = 9/20

The combined area of the pool and deck is 9/20 of the area of the backyard.

Now we subtract the combined area of the pool and deck from the entire area of the backyard.

1 - 9/20 = 20/20 - 9/20 = 11/20

The grassy area is 11/20 of the backyard.

Answer: 11/20

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In a study of the relationship of the shape of a tablet to its dissolution time, 6 disk-shaped ibuprofen tablets and 8 oval-shap
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Answer:

Step-by-step explanation:

This is a test of 2 independent groups. Let μ1 be the mean dissolution time for disk-shaped ibuprofen tablets and μ2 be the mean dissolution time for oval-shaped ibuprofen tablets.

The random variable is μ1 - μ2 = difference in the mean dissolution time for disk-shaped ibuprofen tablets and the mean dissolution time for oval-shaped ibuprofen tablets.

We would set up the hypothesis.

a) The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

For disk shaped,

Mean, x1 = (269.0 + 249.3 + 255.2 + 252.7 + 247.0 + 261.6)/6 = 255.8

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

n1 = 6

Summation(x - mean)² = (269 - 255.8)^2 + (249.3 - 255.8)^2 + (255.2 - 255.8)^2+ (252.7 - 255.8)^2 + (247 - 255.8)^2 + (261.6 - 255.8)^2 = 337.54

Standard deviation, s1 = √(337.54/6) = 7.5

For oval shaped,

Mean, x2 = (268.8 + 260 + 273.5 + 253.9 + 278.5 + 289.4 + 261.6 + 280.2)/8 = 270.7375

n2 = 8

Summation(x - mean)² = (268.8 - 270.7375)^2 + (260 - 270.7375)^2 + (273.5 - 270.7375)^2+ (253.9 - 270.7375)^2 + (278.5 - 270.7375)^2 + (289.4 - 270.7375)^2 + (261.6 - 270.7375)^2 + (280.2 - 270.7375)^2 = 991.75875

Standard deviation, s2 = √(991.75875/8) = 11.1

b) 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)

Therefore,

t = (255.8 - 270.7375)/√(7.5²/6 + 11.1²/8)

t = - 3

c) 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 = [7.5²/6 + 11.1²/8]²/[(1/6 - 1)(7.5²/6)² + (1/8 - 1)(11.1²/8)²] = 613.86/51.46

df = 12

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

p value = 0.011

d) Since alpha, 0.05 > than the p value, 0.011, then we would reject the null hypothesis. Therefore, we can conclude that at 5% significance level, the mean dissolve times differ between the two shapes

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3 years ago
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