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shusha [124]
3 years ago
15

Area of Composed figures PLEASEEE RN

Mathematics
1 answer:
madam [21]3 years ago
5 0

Answer:

33 ft²

Step-by-step explanation:

The area of the outdoor carpet = area of trapezoid + area of rectangle + area of triangle

✔️Area of trapezoid = ½(a + b)h

Where,

a = 6 ft

b = 3 ft

h = 11 - 9 = 2 ft

Area of trapezoid = ½(6 + 3)*2 = 9 ft²

✔️Area of rectangle = L*W = 7 *3 = 21 ft²

✔️Area of triangle = ½bh

b = 2 ft

h = 3 ft

Area = ½*2*3 = 3 ft²

✅Area of the outdoor carpet = 9 + 21 + 3 = 33 ft²

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Step-by-step explanation:

5 liter flask contains 1.25 x 10^8 bacteria.

So, 1liter flask contains = \frac{1.25 \times 10^8}{5} =2.5\times 10^7

Now

0.25 mL contains = 2.5\times 10^7 \times0.25\times 10^{-3} = 6.25\times 10^3

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6 0
3 years ago
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There is sufficient evidence to conclude that, the percentage of the families who own a pet is different than 60%.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomena under the assumption that it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

It Is said That 60% of families own a pet.

Of a sample of 95 families, 70 owned pets.

Whether the percent of families who own pets is different than 60%.

Let P be a proportion of the family who owns pets

Then by the test, we have

H₀: P = 0.60

Hₐ: P ≠ 60

Then by the test statistic, we have

z_o = \dfrac{\bar{P} - P_o}{\sqrt{\dfrac{P_o(1 - P_o)}{n}}}

Where

\bar P  = sample proportion

P₀ = hypothesis proporion

n = sample size

Then we have

\bar P = x/n

\bar P = 70/95

\bar P = 0.74

Then the test statistic will be

z_o = \dfrac{0.74-0.60}{\sqrt{\dfrac{0.60(1-0.60)}{95}}}

z₀ = 2.785

Then the critical region will be

Critical value = ± z_{\alpha /2}

α = 0.05

α/2 = 0.025

Then we have

z₀.₀₂₅ = 1.96

Then the critical value will be

Critical value = ± 1.96

We reject if |z_o| > |z_{\alpha /2}|

We have

|z_o| > |z_{\alpha /2}|\\

2.79 > 1.96

We reject the null hypothesis at a 5% significance level.

There is sufficient evidence to conclude that, the percentage of the families who own a pet is different than 60%.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ4

6 0
2 years ago
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