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3 years ago

5

10 times 3 tens unit and standard form

1 answer:

lorasvet [3.4K]3 years ago

4 0

300, three hundred is the standard form

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Which type of construction is used to construct a triangle's point of balance

aniked [119]

Um do you ave any answer choices ? i would say <span>Pythagorean theorem.
thats my best guess </span>

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4 years ago

Complete the given operation and write the expression in standard form.

Elodia [21]

Step-by-step explanation:

Using synthetic division of polynomials 3x⁴-2x²+4x+9÷x+2. Rewrite the dividend as 3x⁴+0x³-2x²+4x+9

2 | 3 0 -2 4 9

| 6 -12 20 -32

3 -6 10 -16 41

So the quotient is 3x³-6x²+10x-16 with a remainder of 41

Therefore, 3x³-6x²+10x-16+(41/x+2)

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3 years ago

92. Calculate the mean of each data set below. Can you find any shortcuts that allow you to find the mean without having to do m

dimulka [17.4K]

Calculate the mean of each data set below. Can you find any shortcuts that allow you to find the mean without having to do much calculation? Homework help 6, 10, 6, 10 11, 12, 12, 13, 12 0, 5, 4, 8, 0, 7

Answer: To find the mean of the given observations. we just need to first find the sum of the given observations and the divide the calculate sum by the total number of observations.

So here:

Sum of of observations $=6+ 10+ 6+10+11+12+12+13+12+0+ 5+ 4+ 8+ 0+7=116$

Number of observations = 15

Therefore, the mean $=\frac{116}{15}= 7.733$

5 0

4 years ago

Use the formula A=2πrh to find the area of the curved surface of each of the cylinders below. (Express your answers correct to 1

shusha [124]

Answer:

here,

A=2×22÷7×17/2×21

A=22×17×3

A=1122 sq.cm

3 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

4 years ago