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german
4 years ago
8

Tom's age is 5 more than twice mick's age.

Mathematics
2 answers:
xxTIMURxx [149]4 years ago
5 0
The answer is 5+2x because you have to think of micks age as a number
Arturiano [62]4 years ago
3 0
5+2x

5 is Tom’s age plus twice Mick’s age 2c
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Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of
jarptica [38.1K]

Answer:

The "probability that a given score is less than negative 0.84" is  \\ P(z.

Step-by-step explanation:

From the question, we have:

  • The random variable is <em>normally distributed</em> according to a <em>standard normal distribution</em>, that is, a normal distribution with \\ \mu = 0 and \\ \sigma = 1.
  • We are provided with a <em>z-score</em> of -0.84 or \\ z = -0.84.

Preliminaries

A z-score is a standardized value, i.e., one that we can obtain using the next formula:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

  • <em>x</em> is the <em>raw value</em> coming from a normal distribution that we want to standardize.
  • And we already know that \\ \mu and \\ \sigma are the mean and the standard deviation, respectively, of the <em>normal distribution</em>.

A <em>z-score</em> represents the <em>distance</em> from \\ \mu in <em>standard deviations</em> units. When the value for z is <em>negative</em>, it "tells us" that the raw score is <em>below</em> \\ \mu. Conversely, when the z-score is <em>positive</em>, the standardized raw score, <em>x</em>, is <em>above</em> the mean, \\ \mu.

Solving the question

We already know that \\ z = -0.84 or that the standardized value for a raw score, <em>x</em>, is <em>below</em> \\ \mu in <em>0.84 standard deviations</em>.

The values for probabilities of the <em>standard normal distribution</em> are tabulated in the <em>standard normal table, </em>which is available in Statistics books or on the Internet and is generally in <em>cumulative probabilities</em> from <em>negative infinity</em>, - \\ \infty, to the z-score of interest.

Well, to solve the question, we need to consult the <em>standard normal table </em>for \\ z = -0.84. For this:

  • Find the <em>cumulative standard normal table.</em>
  • In the first column of the table, use -0.8 as an entry.
  • Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
  • The intersection of these two numbers "gives us" the cumulative probability for z or \\ P(z.

Therefore, we obtain \\ P(z for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".

This represent the area under the <em>standard normal distribution</em>, \\ N(0,1), at the <em>left</em> of <em>z = -0.84</em>.

To "draw a sketch of the region", we need to draw a normal distribution <em>(symmetrical bell-shaped distribution)</em>, with mean that equals 0 at the middle of the distribution, \\ \mu = 0, and a standard deviation that equals 1, \\ \sigma = 1.

Then, divide the abscissas axis (horizontal axis) into <em>equal parts</em> of <em>one standard deviation</em> from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).  

Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, \\ -\sigma, from it). All the area to the left of this value must be shaded because it represents \\ P(z and that is it.

The below graph shows the shaded area (in blue) for \\ P(z for \\ N(0,1).

7 0
3 years ago
Someone tell me the anwser and i need to show my work too
Zinaida [17]
First you need to find how much 20% of $18.50 is. To do this, you need to multiply 18.50 by 20% which in decimal form is 0.2
$18.50x0.2=$3.70
Now you need to subtract $3.70 from $18.50
$18.50-$3.70=$14.80
Now that you have the new price, you need to find 6.75% of it to get the sales tax, so you multiply $14.80 by 6.75%, which is 0.0675 as a decimal.
$14.80x0.0675=$0.999 or $1 
Now you add the sales tax to $14.80
$14.80+$1=$15.80 or $14.80+$0.999=$15.799
6 0
3 years ago
Read 2 more answers
What is 1.16 as afraction
saul85 [17]

Answer:

23.2/20

Step-by-step explanation:

4 0
3 years ago
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A corporate team-building event costs $19plus an additional $1 per attendee. How many attendees can there be, at most, if the bu
Mars2501 [29]

Answer:

There can be at most 12 attendees in a corporate team-building event.

Step-by-step explanation:

Let x denotes number of attendees in a corporate team-building event.

Fixed cost = $19

Cost charged per attendee = $1

Budget for the corporate team-building event = $31

Therefore,

19+1(x)\leq 31\\19+x\leq 31\\x\leq 31-19\\x\leq 12

So, there can be at most 12 attendees in a corporate team-building event.

8 0
3 years ago
A group of friends is sharing 2 1/2 pounds of berries if 5 friends are sharing berries how many pounds of berries does each frie
Korvikt [17]
1/2 pounds,


assuming that each friend takes an equal part of the 2 1/2 pounds, then 2 1/2 divided by 5 would get you 1/2 pound
4 0
3 years ago
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