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3 years ago

Which two numbers does lie between on a number line?

Mathematics

2 answers:

mina [271]3 years ago

8 0

HI!

So the square root of 128 is 11.31.

You would know that it is more than 11 since 11^2 is 121 which is 7 less than 128.

So the answer would be between 11.3 and 11.4

also 12^2 is 144

121 and 144 are 23 points away and that divided by 10 is 23/10. 23/10 is a little more than 2. so everytime 121 is added by 2, go up by .1

so....

121+2 = 123

11 + .1 = 11.1

123+ 2 = 125

11.1 + .1 = 11.2

125+2=127

11.2+.1=11.3

127+2=129

11.3+.1=11.4

and you would know that square root of 128 is between 11.3 and 11.4

Hope this helps!

allsm [11]3 years ago

5 0

√128 = 11.31

11.31 > 11.3 and 11.31 < 11.4

Answer

11.3 and 11.4

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3 years ago

Use words to write a comparison statement for the problem above.

pantera1 [17]

Looking at the question we can see that it says:

$41253>35214$

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We can say that 2 is less than 5 so in a symbolic form it can be written as,

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Similarly,

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We can say that 7 is greater than 5, in a symbolic form it can be written as,

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3 years ago

Read 2 more answers

The diagram shows 5cm×5cm×5cm cube calculate the length of the diagonal ab

mario62 [17]

Answer:

AB = $5\sqrt{3}$ cm

Step-by-step explanation:

As we can see from the figure, BCDE is a square with each corner equal to 90°.

So that, BDE is a right triangle with corner BED equal to 90°

As BDE is a right triangle, according to Pythagoras theorem, we have:

$BD^{2} = BE^{2} +ED^{2} = 5^{2} + 5^{2} = 25 + 25 = 50$ cm

As the diagram s the cube, so that it can be seen that AD is perpendicular to the surface BCDE

=> AD is perpendicular to BD

=> ADB is the right triangle with corner ADB equal to 90°.

As ADB is the right triangle, ccording to Pythagoras theorem, we have:

$AB^{2} =AD^{2} + BD^{2} = 5^{2} +50 = 25 +50 = 75$ cm

=> $AB = \sqrt{75} = 5\sqrt{3}$ cm

Conclusion: AB = $5\sqrt{3}$ cm

4 0

3 years ago

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mea

QveST [7]

A.
$\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)$

Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.

$\mathbb P(Z>-2)=\mathbb P(|Z|2)=0.95+0.025=0.975$

b.
$\mathbb P(1100$

About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.

c. You're looking for $k$ such that

$\mathbb P(X>k)=0.10$

Since

$\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10$

occurs for $k^*\approx1.2816$ , it follows that

$\dfrac{k-1200}{100}=1.2816\implies k\approx1328$

So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.

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4 years ago