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3 years ago

Determine the length of the longest piece of timber which could be stored in a rectangular shed 6cm by 5cm by 2m high

Mathematics

2 answers:

Brut [27]3 years ago

6 0

Answer:

6 90u7 jhi

Step-by-step explanation:

Sladkaya [172]3 years ago

4 0

Answer:

200.2

Step-by-step explanation:

I would hate to own such a shed. It's not very big.

2 meters = 2 * 100 cm = 200 cm

The diagonal of the floor is the diagonal of 6cm by 5cm. Using a^2 + b^2 = c^2

5^2 + 6^2 = 25 + 36 = 61

The square root of 61 is 7.81 cm

Now the diagonal of the floor diagonal and the height

a^2 + h^2 = board^2

61 + 200^2 = board^2

61 + 40000 = board^2

40061 = board^2

sqrt(board^2) = sqrt(40061)

board = 200.2cm long.

Maybe you should check the question for units.

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String a is 35 centimeters long. String b is 5 times long as string a. Both are necessary to create a decorative bottle. Find th

vladimir2022 [97]

String a is 35 cm.....string b is 5 times as long as string a.....
so string b is 5(35) = 175 cm....both are needed....
175 + 35 = 210 cm for 1 bottle

for 17 bottles...
17 * 210 = 3570 cm <== 3570 cm are needed for 17 bottles

6 0

3 years ago

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Plz answer i suck at math :)

MatroZZZ [7]

Answer:

4.08

Step-by-step explanation:

26.08 +(-6)²*(3.4-8.9)÷9

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26.08-198÷9

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3 years ago

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Solve the inequality, -3 < n -8

vesna_86 [32]

Add 8 on both sides in order to isolate n:

-3 + 8 < n - 8 + 8

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8 0

3 years ago

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

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3 years ago

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denis23 [38]

Answer:

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3 years ago

Determine the length of the longest piece of timber which could be stored in a rectangular shed 6cm by 5cm by 2m high​

Determine the length of the longest piece of timber which could be stored in a rectangular shed 6cm by 5cm by 2m high