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3 years ago

What is the length of segment XY?

Mathematics

2 answers:

Alexandra [31]3 years ago

7 0

Answer: Third option.

Step-by-step explanation:

You need to use the formula for calculate the distance between two points. This is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

You can observe in the graph that the coordinates of the point X and the point Y are the following:

X(-4,0) and Y(3,2)

Knowing this, you can substitute the coordinates into the formula.

You get that the lenght of the segment XY is:

$d_{(XY)}=\sqrt{(3-(-4))^2+(2-0)^2}\\\\d_{(XY)}=\sqrt{53}\ units$

This matches with the third option.

SashulF [63]3 years ago

5 0

Answer:

7.28 units to the nearest hundredth.

Step-by-step explanation:

Use the Pythagoras theorem.

If you examine the graph you see that the line segment is the hypotenuse of a right triangle with legs of length 2 and 7.

XY^2 = 2^2 + 7^2

XY^2 = 53

XY = √53

XY = 7.28.

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The amount of soft drink in a bottle is a Normal random variable. Suppose that in 7% of the bottles containing this soft drink t

WITCHER [35]

Answer:

The mean is 15.93 ounces and the standard deviation is 0.29 ounces.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

7% of the bottles containing this soft drink there are less than 15.5 ounces

This means that when X = 15.5, Z has a pvalue of 0.07. So when X = 15.5, Z = -1.475.

$Z = \frac{X - \mu}{\sigma}$

$-1.475 = \frac{15.5 - \mu}{\sigma}$

$15.5 - \mu = -1.475\sigma$

$\mu = 15.5 + 1.475\sigma$

10% of them there are more than 16.3 ounces.

This means that when X = 16.3, Z has a pvalue of 1-0.1 = 0.9. So when X = 16.3, Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{16.3 - \mu}{\sigma}$

$16.3 - \mu = 1.28\sigma$

$\mu = 16.3 – 1.28\sigma$

From above

$\mu = 15.5 + 1.475\sigma$

$15.5 + 1.475\sigma = 16.3 – 1.28\sigma$

$2.755\sigma = 0.8$

$\sigma = \frac{0.8}{2.755}$

$\sigma = 0.29$

The mean is

$\mu = 15.5 + 1.475\sigma = 15.5 + 1.475*0.29 = 15.93$

The mean is 15.93 ounces and the standard deviation is 0.29 ounces.

6 0

3 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

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Answer:

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Step-by-step explanation:

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Priya has completed 9 exam questions. This is 60% of the questions on the exam. How many questions are on the exam?

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3 years ago

What is the length of segment XY?​

What is the length of segment XY?