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nlexa [21]
3 years ago
11

Nine twentieths is equal

Mathematics
2 answers:
ruslelena [56]3 years ago
5 0
An equivalent fractions would be 18/40.
kondaur [170]3 years ago
4 0
Nine twentieths is equal to...

9/20 and 0.45.
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What is the unit rate of 15 and 2
Katena32 [7]
Https://www.khanacademy.org/math/in-sixth-grade-math/ratio-and-proportion/unitary-method/v/finding-unit-rates
4 0
3 years ago
PLZ HELP ME
NeTakaya

Answer:

  see below

Step-by-step explanation:

The shading is below the line in both cases. The dashed line has negative slope, so the inequality associated with it is ...

  y < -2x +4 . . . . . . note that the < symbol has no "or equal to"

The solid line has positive slope, so its inequality is ...

  y ≤ 2x +4

Both these inequalities are found in choice B.

6 0
2 years ago
There are 75 students enrolled in camp. The day before the camp begins, 8℅ of the students cancel. how many students actually at
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69 kids still attend
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3 years ago
Find parametric equations through point P=(2,2,7) in the direction of the vector v = (44, 14, -20)?
ycow [4]

9514 1404 393

Answer:

  (x, y, z) = (2+44t, 2+14t, 7-20t)

Step-by-step explanation:

One way to write parametric equations for line L is ...

  L = P + t·<em>v</em>

where P is the given point and <em>v</em> is the given direction vector. Using that form, we have ...

  (x, y, z) = (2+44t, 2+14t, 7-20t)

__

If you like, you can remove a common factor of 2 from the coefficients of t.

  (x, y, z) = (2+22t, 2+7t, 7-10t)

5 0
2 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
3 years ago
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