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3 years ago

a playground is 750 M long and 250 m broad find the cost of levelling at rupees 16 per 100 square metres

Mathematics

1 answer:

barxatty [35]3 years ago

4 0

Answer:

Area of play ground = Length × Breadth = 750 × 250 = 187500. a Cost of levelling the ground = 187500 × 16 100 = Rs . 30000

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in the backwaters of Kerala a boatman was throwing a Boat using a long pole of length 4.80 when he put the pole vertically into

Hitman42 [59]

Answer:

3.60m

Step-by-step explanation:

In backwaters of Kerala, a boatman was throwing a boat with a long pole which has a length of 4.80m

The boatman put the pole into the water vertically, the part of the pole that was seen above the water level is 0.25m

The first step is to calculate the length of the pole that is seen above the water level

= 4.80 × 0.25

= 1.20m

Therefore the depth of the water at that point can be calculated as follows

=4.80m - 1.20m

= 3.60m

Hence the depth of the water at that point is 3.60m

4 0

3 years ago

I need help. This is due later today.

REY [17]

Answer:

D is one of them

Step-by-step explanation:

4 0

3 years ago

A manufacturing company uses two different machines, A and B, each of which produces a certain item part. The number of defectiv

damaskus [11]

Answer:A-The mean is 0 and the distribution is approximately normal.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please help. I’ll mark you as brainliest if correct!

djverab [1.8K]

Answer:

Step-by-step explanation:

children=c

adults=a

c+a=359

a=359-c

2.75c+6a=1621

2.75 c+6(359-c)=1621

2.75 c+2154-6c=1621

-3.25 c=1621-2154

-3.25 c=-533

$-\frac{325}{100} c=-533\\-\frac{13}{4} c=-533\\c=-533 \times \frac{-4}{13} =41 \times 4=164 \\children=164\\adults=359-164=195$

7 0

3 years ago

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.

3 0

3 years ago

a playground is 750 M long and 250 m broad find the cost of levelling at rupees 16 per 100 square metres​

a playground is 750 M long and 250 m broad find the cost of levelling at rupees 16 per 100 square metres