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3 years ago

Describing How to Format Tables Using Table and Border Styles

Mathematics

2 answers:

ivann1987 [24]3 years ago

5 0

Answer:

He clicked the design tab

He selected the whole table

He changed the table style

Step-by-step explanation:

Just did the assignment :)

andrezito [222]3 years ago

5 0

Answer:

1, 2, 3

Step-by-step explanation:

edge 2020

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

7 0

4 years ago

on her last 3 terms in algebra, Myka had scored a 83,87, and a 92. What score does she need to get on her text test if she wants

Ugo [173]

At least a 98 or it won’t be above a 90

8 0

3 years ago

The diagram shows a convex polygon <br><br> solve for b

ikadub [295]

Just add all of them up the sides

8 0

3 years ago

Anne plans to increase the prices of all the items in her store by 5%. To the nearest cent, how much will the artist save if the

Advocard [28]

5% + 5% = 10%
He will save 10%

8 0

4 years ago

Read 2 more answers

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0

3 years ago