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Sliva [168]
2 years ago
5

An isosceles triangle has base 6 and height 11. find the maximum possible area of a rectangle that can be placed inside the tria

ngle with one side on the base of the triangle. (hint: similar triangles may be of assistance.)
Mathematics
1 answer:
svetoff [14.1K]2 years ago
5 0
The area of the triangle is 33
You might be interested in
2.5 kg of cabbages cost £1.20<br><br> Work out the cost of 3.75 kg
s2008m [1.1K]
Thank you for your question!

In order to get the cost of 1 kg of cabbage, divide the cost by the quantity:

1.20/2.5
0.48

Now, multiply the amount for 1 kg to 3.75, because you need 3.75 times the amount.

0.48 * 3.75
1.80

So, it would cost 1.80 pounds for 3.75 kg.
Hope this helped!


8 0
2 years ago
There are nine teddy bears in the diagram below. How many TOTAL squares are<br> there?
IgorC [24]

Answer: 14

Step-by-step explanation: 9 small squares with one teddy bear in each, 4 medium squares with 4 bears in each medium square, and 1 big square for the entire thing.

6 0
2 years ago
Read 2 more answers
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:

  1. Use the <em>cumulative</em> <em>standard normal distribution</em> to find the probability.
  2. 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.
  3. 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
Given that a rectangle with length $3x$ inches and width $x + 5$ inches has the property that its area and perimeter have equal
scoray [572]

Answer:

First, we need to know how to calculate the area and the permiter of a rectangle.

To calculate the area, we multiply base by height and to calculate the perimeter, we sum all sides.

Knowing this, we can say that the area is 3x * (x+5) and the perimiter is 3x + 3x + x + 5 + x + 5, as we know both are the same, we write it as an equation:

3x * (x+5) = 3x + 3x + x + 5 + x + 5

Now we solve the equation:

3x^2 +15x = 6x + 2x + 10

3x^2+15x =8x + 10

3x^2+15x-8x-10=0

3x^2+7x-10=0\\\\x_1=\frac{-10}{3}\\x_2 = 1

As the negative result doesn't have sense, we only pick the second one: 1.

If x = 1, then area would be 3*6 = 18 square inches and perimeter 3+3+6+6 = 18 inches

7 0
1 year ago
Read 2 more answers
A team of dermatological researchers were studying skin cancer among 60-70 year old American men. Suppose you were a part of the
Lostsunrise [7]

Answer:

The correct answer is Option A.) The standard deviation of the distribution of sample means is less than the population standard deviation of the number of cancer spots.

Step-by-step explanation:

The standard deviation of the distribution of sample means is less than the population standard deviation of the number of cancer spots.

3 0
1 year ago
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