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4 years ago

What is the distance between (6, 4) and (8, -5)?

Mathematics

1 answer:

MrRissso [65]4 years ago

6 0

Answer:

9.21 units

Step-by-step explanation:

Let d be the distance between given points.

$\therefore \: d = \sqrt{ {(6 - 8)}^{2} + { \{4 - ( - 5) \}}^{2} } \\ \hspace{16 pt}= \sqrt{ {( - 2)}^{2} + { \{4 + 5 \}}^{2} } \\\hspace{16 pt} = \sqrt{ {( - 2)}^{2} + { \{9 \}}^{2} } \\ \hspace{16 pt}= \sqrt{ 4 + 81} \\\hspace{16 pt} = \sqrt{ 85} \\ \hspace{16 pt}= 9.21954446 \\ \: \: \: \: \: \purple{ \bold{d = 9.21 \: units }}\\ \\$

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If $1000 is invested at 8% interest, compounded annually, then after n years the investment is worthan = 1000(1.08)ndollars.(a)

Luba_88 [7]

Answer:

Diverge

Step-by-step explanation:

(a)

1st year: $1000 * 1.08^1 = \$1080$

2nd year: $1000 * 1.08^2 = \$1166.4$

3rd year: $1000 * 1.08^3 = \$1259.71$

4th year: $1000 * 1.08^4 = \$1360.49$

5th year: $1000 * 1.08^5 = \$1469.33$

(b) The sequence is divergent, because if we take the derivative of the function with respect to n year:

$(1000*1.08^n)^{'} = 1000ln(n)*1.08^n$

This is a positive, meaning the slope of the function is positive. If we take the second derivative using product rule

$(1000ln(n)*1.08^n)^{'} = 1000\frac{ln(n)}{n}1.08^n$

This is also positive when n > 0. Therefore, the slope is positive and increasing. This means the sequence diverges.

5 0

3 years ago

Write the following number in standard notation: 6.52 × 103.

solniwko [45]

6520 is the answer.

6.52 * 10^3 === 6.52 * 1000 = 6520

6 0

4 years ago

Read 2 more answers

2. In a school of 320 students, 85 students are in the band, 200 students are on sports teams, and 60 students

Alja [10]

Answer:

60 students are involved in either band or sports.

95 students are not involved in either sports or band.

Step-by-step explanation:

Given:

In a school of 320 students, 85 students are in the band, 200 students are on sports teams, and 60 students participate in both activities.

Now, to find the students are involved in either sports or band. And students who are not involved in either sports or band.

Total number of students = 320.

Number of students in the band = 85.

Number of students in the sports team = 200.

Number of students participate in both activities = 60.

Thus, 60 students are involved in either band or sports.

So, the number of students in the band adding number of students in the sports team we get:

200 + 85 = 285.

Then, to subtract the number of students participate in both activities we get:

285 - 60 = 225.

Thus, the remaining students are = 225.

Now, to get the students who are not involved in either sports or band we subtract the total number of students from the remaining students:

$320-225$

$=95.$

Therefore, 60 students are not involved in either band or sports. 95 students are not involved in either sports or band.

8 0

3 years ago

Can someone help me please ? It would be greatly appreciated:-)

WITCHER [35]

I think it might be a ? it sounds the most reasonable .

7 0

3 years ago

4. The distribution of blood cholesterol level in the population of young men aged 20 to 34 years is close to Normal, with mean

Pie

Answer:

a) 38.59% probability that a young man (aged 20 to 34) has a cholesterol level greater than 200 milligrams per deciliter.

b) By the Central Limit Theorem, the mean of the distribution of the sample mean would be 188 milligrams per deciliter

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.

In this problem, we have that:

$\mu = 188, \sigma = 41$

a. Find the probability that a young man (aged 20 to 34) has a cholesterol level greater than 200 milligrams per deciliter.

This is 1 subtracted by the pvalue of Z when X = 200. So

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

$Z = \frac{200 - 188}{41}$

$Z = 0.29$

$Z = 0.29$ has a pvalue of 0.6141

1 - 0.6141 = 0.3859

38.59% probability that a young man (aged 20 to 34) has a cholesterol level greater than 200 milligrams per deciliter.

b. Suppose you measure the cholesterol level of 100 young men chosen at random and calculate the sample mean. If you did this many times, i. what would be the mean of the distribution of the sample mean

By the Central Limit Theorem, the mean of the distribution of the sample mean would be 188 milligrams per deciliter

5 0

3 years ago