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jok3333 [9.3K]
3 years ago
15

The figure shows a carpeted room. How many square feet of the room are carpeted?

Mathematics
1 answer:
Bezzdna [24]3 years ago
4 0
Ah okay, now I can answer this. 

Take separate areas. The bottom section has an area of 20 feet, while the top rectangle has an area of 24 feet. 

The answer is B. 
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[PLEASE BE QUICK] Write the equation of the line in point-slope form that passes through the given point with the given slope. 5
ladessa [460]

Answer:

y-5=-3(x-1)           simplified: y=-3x+8

Step-by-step explanation:

y-y1=m(x-x1)

plug in point and slope given

y-5=-3(x-1)

y=-3x+3+5

y=-3x+8

3 0
3 years ago
Suppose that a study of elementary school students reports that the mean age at which children begin reading is 5.9 years with a
Bas_tet [7]

Answer:

5.9 years.

Step-by-step explanation:

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

Mean of the population is \mu = 5.9

If a sampling distribution is created using samples of the ages at which 69 children begin reading, what would be the mean of the sampling distribution of sample means?

By the Central Limit Theorem, the same population mean, of 5.9 years.

3 0
3 years ago
What divided by 5 equal 1.21
oksano4ka [1.4K]
Set the number as x
x/5=1.21
Multiply by 5 on each side to cancel the 5 in the denominator.
x=6.05

Check your answer:
6.05/5=1.21
1.21=1.21
7 0
4 years ago
C) Three yam tubers are chosen at random from 15 tubers of which 5 are spoilt. Find the probability that, of the three chosen tu
Dmitry_Shevchenko [17]

\displaystyle\\|\Omega|=\binom{15}{3}=\dfrac{15!}{3!12!}=\dfrac{13\cdot14\cdot15}{2\cdot3}=455

a)

\displaystyle\\|A|=\binom{10}{3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{2\cdot3}=120\\\\P(A)=\dfrac{120}{455}=\dfrac{24}{91}\approx26.4\%

b)

\displaystyle\\|A|=\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdot5}{2}=10\\\\P(A)=\dfrac{10}{455}=\dfrac{2}{91}\approx2.2\%

c)

\displaystyle\\|A|=\binom{10}{2}\cdot5=\dfrac{10!}{2!8!}\cdot5=\dfrac{9\cdot10}{2}\cdot5=225\\\\P(A)=\dfrac{225}{455}=\dfrac{45}{91}\approx49.5\%

d)

A - at least one is spoilt

A' - none is spoilt

P(A)=1-P(A')

We calculated P(A') in a).

Therefore

P(A)=1-\dfrac{24}{91}=\dfrac{67}{91}\approx73.6\%

7 0
2 years ago
The composite figure is made up of a <br> two<br><br> figures<br> on top of a <br><br> .
mars1129 [50]

That’s not a question? I need you to specify on what the question is then I can edit this post and let you know the answer

3 0
3 years ago
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