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musickatia [10]

2 years ago

9

At the farmers market, Nathaniel buys two dozen oranges at $3.50 per dozen, four kilograms of apples at $3.25 per kilogram, two

kilograms of carrots at $1.90 per kilogram, and one kilogram of potatoes for $1.20. He received a 5% discount off his total bill. How much money did Nathaniel spend on fruits and vegetables at the farmers market?

1 answer:

irga5000 [103]2 years ago

3 0

Oranges
2(3.50) = 7.00
apples
4(3.25) = 13.00
carrots
2(1.90) = 3.80
potatoes
1(1.20) = 1.20
7 + 13 + 3.80 + 1.20 = 25.00 * .05 = 1.25
$25.00 - 1.25 = $23.75

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Answer:

Austin's hourly wage is $8.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

Tara's hourly wage is x.

Kayte's hourly wage is y.

Austin's hourly wage is z.

Tara earns twice as much per hour as Kayte.

This means that $x = 2y$

Kayte earns $3 more per hour than Austin.

This means that $y = z + 3$

As a group, they earn $41 per hour.

This means that $x + y + z = 41$

What is Austin's hourly wage?

This is z.

$x + y + z = 41$

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$x + y + z = 41$

$2z + 6 + z + 3 + z = 41$

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$z = 8$

Austin's hourly wage is $8.

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The answer is a.3/4ft

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Answer:

Alison wins against Kevin by 0.93 s

Step-by-step explanation:

Alison covers the last 1/4 of the distance in 3 seconds, at a constant acceleration $a_a$ , we have the following equation of motion

$s/4 = a_at_a^2/2$

where s (m) is the total distance, ta = 3 s is the time

$s = 4a_a3^2/2 = 18a_a$

$a_a = s/18$

Similarly, Kevin overs the last 1/3 of the distance in 4 seconds, at a constant acceleration $a_k$ , we have the following equation of motion:

$s/3 = a_kt_k^2/2$

tk = 4 s is the time

$s = 3a_k4^2/2 = 24a_k$

$a_k = s/24$

Since $a_a = s/18$ we can conclude that $a_a > a_k$ , so Alison would win.

The time it takes for Alison to cover the entire track

$s = a_aT_a^2/2$

$T_a^2 = 2s/a_a = 2s/(s/18) = 36$

$T_a = \sqrt{36} = 6 s$

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2 years ago

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Step-by-step explanation:

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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.

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2 years ago