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Vedmedyk [2.9K]

2 years ago

13

Tiny's Tattoo Parlor sold 389 tattoos this month. 43 of those tattoos were arm tattoos. Based on this data, what is a reasonable

estimate of the probability that the next customer does not get an arm tattoo?

2 answers:

antiseptic1488 [7]2 years ago

8 0

Answer
(389-43)/389=346/389

OverLord2011 [107]2 years ago

5 0

Answer:

346/389

Step-by-step explanation:

correct on khan academy!

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A fruit stand sells mangoes for $3.49 per pound, papayas for $1.40 per pound, and coconuts for $1.24 per pound. Part A Which exp

Oksana_A [137]

Part A is C. because it is $3.49 per mango, so it is $3.49m, and $1.40 per papaya, so that is $1.40p, and $1.24 per coconut, so that is $1.24c. You add those up and the expression would be 3.49m + 1.40p + 1.24c.

Part B is B.

Because 3 times 3.49 is 10.47, which is the cost for the mangoes. And 1.4 times 4 is 5.6, which is the cost for the papayas. And 6 times 1.24 is 7.44, which is the cost of the coconuts. Add 10.47, 5.6, and 7.44 and you get 23.51 which is the answer.

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2 years ago

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With reference to the figure match the angles and arcs to their measures

photoshop1234 [79]

Angle BAC = (180 - 114) degrees = 66 degrees [angle on a straight line]

Angle OCA = Angle OBA = 90 degrees    [angle at the point where the tangent and the radius meet]

Thus, the measure of arc BC = (360 - 66 - 90 - 90) degrees = 114 degrees [sum of interior angles of a quadrilateral]

Angle CDE = (180 - 124) degrees = 56 degrees [angle on a straight line]

Angle OCD = Angle OED = 90 degrees    [angle at the point where the tangent and the radius meet]

Thus, the measure of arc CE = (360 - 56 - 90 - 90) degrees = 124 degrees    [sum of interior angles of a quadrilateral]

Given that the measure of arc BC is 114 degrees and the measure of arc CE is 124 degrees, thus the measure of arc BE = (360 - 114 - 124) degrees = 122 degrees [angle at a point]

Angle OBF = Angle OEF = 90 degrees    [angle at the point where the tangent and the radius meet]

Thus, angle BFE = (360 - 122 - 90 - 90) degrees = 58 degrees.

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2 years ago

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Torrey starts a new job with an annual salary of $60,000. For each year she continues to work for the same company, she will rec

GarryVolchara [31]

The third choice will work

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2 years ago

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The number of newly reported crime cases in a county in New York State is shown in the accompanying table, where x represents th

Novay_Z [31]

Answer:

Step-by-step explanation:

I use 84+ CE

stat edit, then fill in the #s

then

vars 5

then

2'nd stat plot, on

then, click stat

Click arrow 1 time to the left to get to Calc

then click (4)(LinReg(ax+b))

then click enter 5 times

(y=-25.31428571x+1000.285714

y=-25.3x+1000.3

now, lets use computer:

y=-25.31(543)+1000.3

y=-12743.03

round to the biggest whole number )

this doesn't really work, so I will put 1999, 2000, 2001, 2002, 2003, 2004 instead of 0, 1, 2, 3, 4, 5 and do the same thing

now I get

y=-25.31428571x+51603.54286

y=-25.3x+51603.5

now, lets use computer:

y=-25.3(543)+51603.5

y=37865.6

round to the biggest whole number:

y=37866

so, year 37866

7 0

2 years ago

g Assume that the distribution of time spent on leisure activities by adults living in household with no young children is norma

OLga [1]

Answer:

"<em>The probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day</em>" is about 0.8749.

Step-by-step explanation:

We have here a <em>random variable</em> that is <em>normally distributed</em>, namely, <em>the</em> <em>time spent on leisure activities by adults living in a household with no young children</em>.

The normal distribution is determined by <em>two parameters</em>: <em>the population mean,</em> $\\ \mu$ , and <em>the population standard deviation,</em> $\\ \sigma$ . In this case, the variable follows a normal distribution with parameters $\\ \mu = 4.5$ hours per day and $\\ \sigma = 1.3$ hours per day.

We can solve this question following the next strategy:

Use the <em>cumulative</em> <em>standard normal distribution</em> to find the probability.
Find the <em>z-score</em> for the <em>raw score</em> given in the question, that is, <em>x</em> = 6 hours per day.
With the <em>z-score </em>at hand, we can find this probability using a table with the values for the <em>cumulative standard normal distribution</em>. This table is called the <em>standard normal table</em>, and it is available on the Internet or in any Statistics books. Of course, we can also find these probabilities using statistics software or spreadsheets.

We use the <em>standard normal distribution </em>because we can "transform" any raw score into <em>standardized values</em>, which represent distances from the population mean in standard deviations units, where a <em>positive value</em> indicates that the value is <em>above</em> the mean and a <em>negative value</em> that the value is <em>below</em> it. A <em>standard normal distribution</em> has $\\ \mu = 0$ and $\\ \sigma = 1$ .

The formula for the <em>z-scores</em> is as follows

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

Solving the question

Using all the previous information and using formula [1], we have

<em>x</em> = 6 hours per day (the raw score).

$\\ \mu = 4.5$ hours per day.

$\\ \sigma = 1.3$ hours per day.

Then (without using units)

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{6 - 4.5}{1.3}$

$\\ z = \frac{1.5}{1.3}$

$\\ z = 1.15384 \approx 1.15$

We round the value of <em>z</em> to two decimals since most standard normal tables only have two decimals for z.

We can observe that z = 1.15, and it tells us that the value is 1.15 standard deviations units above the mean.

With this value for <em>z</em>, we can consult the <em>cumulative standard normal table</em>, and for this z = 1.15, we have a cumulative probability of 0.8749. That is, this table gives us P(z<1.15).

We can describe the procedure of finding this probability in the next way: At the left of the table, we have z = 1.1; we can follow the first line on the table until we find 0.05. With these two values, we can determine the probability obtained above, P(z<1.15) = 0.8749.

Notice that the probability for the z-score, P(z<1.15), of the raw score, P(x<6) are practically the same, $\\ P(z$ . For an exact probability, we have to use a z-score = 1.15384 (without rounding), that is, $\\ P(z$ . However, the probability is approximated since we have to round z = 1.15384 to z = 1.15 because of the use of the table.

Therefore, "<em>the probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day</em>" is about 0.8749.

We can see this result in the graphs below. First, for P(x<6) in $\\ N(4.5, 1.3)$ (red area), and second, using the standard normal distribution ( $\\ N(0, 1)$ ), for P(z<1.15), which corresponds with the blue shaded area.

5 0

2 years ago