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3 years ago

What is the distance between (-5,0) and (-5,-12)

Mathematics

2 answers:

11111nata11111 [884]3 years ago

8 0

The distance is 12.

BaLLatris [955]3 years ago

4 0

$\text {Distance = } \sqrt{(0+12)^2+(-5+5)^2} = \sqrt{12^2} = 12$

Answer: The distance is 12 units.

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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

3 years ago

Let the sample space be Upper S equals StartSet 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 comma 9 comma 10 EndSe

Marina86 [1]

Answer:

Probability of event E = (4/10) = 0.40

Step-by-step explanation:

Note that n(E) is the number of outcomes in E

n(S) is the number of outcomes in S

And with each outcome having equal likelihood of occurring,

SetS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

n(S) = 10

Event E = {1, 4, 7, 9}.

n(E) = 4

Probability of event E = n(E)/n(S)

Probability of event E = (4/10) = 0.40

Hope this Helps!!

4 0

3 years ago

Please help solve!!!!!!!!!!

Deffense [45]

Answer:

Height = 17.31821 feet

Step-by-step explanation:

Explanation

From graph

Given angle θ = 6.5° and a = 152 feet

we need to find the height (h)

By using the tangent rule

$tan \alpha = \frac{opposite side}{adjacent side}$

$tan 6.5 = \frac{b}{152}$

Cross multiplication , we get

b = 152 × tan 6.5°

b = 152 × 0.113935

b = 17.31821 feet

Height = 17.31821 feet

6 0

3 years ago

Here is an array of ten integers 6 4 0 3 9 8 1 7 2 5

garik1379 [7]

Answer:

$[4,0,3,1,2,5]$ and $[6,9,8,7]$

Step-by-step explanation:

GIVEN: an array of ten integers $6,4,0,3,9,8,1,7,2,5$ .

TO FIND: If we partition this array using Quick sort's partition function and using $5$ for the pivot. List the elements of the resulting array after the partition finishes.

SOLUTION:

quick sort is a divide and conquer algorithm in which an array is partitioned into sub-arrays about an pivot element by checking whether elements are greater than pivot or and then sub arrays are sorted recursively.

Here $5$ is the pivot element.

two arrays will be created, in first array element less than or equal to pivot element are stored in other elements greater than pivot element are stored.

Starting from first element of array

elements in first array will be $=[4,0,3,1,2,5]$

elements in second array will be $=[6,9,8,7]$

Hence the resulting array after the partition finishes are $[4,0,3,1,2,5]$ and $[6,9,8,7]$

6 0

3 years ago

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

1 year ago