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9966 [12]
3 years ago
6

PLEASE HURRY Figure ABCD is transformed to figure A'B'C'D':

Mathematics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

∠D'A'B'

Step-by-step explanation:

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1) The cost of constructing a path 5 m broad inside the boundary of a square lawn at Rs 36.25 per sq. metre is Rs 90625. What is
Svetach [21]

Let , side lawn is a.

Area of boundary is :

A=4\times5\times a\\\\A=20a\ m^2

Now, total cost is given by :

T=20a\times 36.25\\\\90625=725a\\\\a=125\ m

Area left is :

A'=a^2-20a\\\\A'=125^2-20\times 125\ m^2\\\\A'=13125\ m^2

Price to empty space with turfs is :

P=A'\times 20\\\\P=13125\times 20\\\\P=262500

Therefore, the cost of covering the empty space with turfs at the rate of Rs 20 per sq. metre is Rs 262500.

Hence, this is the required solution.

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3 years ago
If the simple inttrest on $ 2.000 for 2 years then what is the intrest rate
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The inttrest rate would be 4
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3 years ago
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Anna71 [15]

Answer:

D. 44 cubic inches

Step-by-step explanation:

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5 0
3 years ago
What is the slope of the line that passes through the points (4, 6) and (−16,−18)? Write your answer in simplest form.
iragen [17]

Step-by-step explanation:

Slope of the line: 6/5.

6 0
3 years ago
Read 2 more answers
It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

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We need a sample size of at least 75.

6 0
3 years ago
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