18182737756l748484884+1727727123444484747747474747477474747474747

I am learning to make meme what u think?

SCORPION-xisa [38]

It’s great. Can I do all your homework?

3 0

2 years ago

Read 2 more answers

Please I need help!!!

Valentin [98]

Answer:

Solution given:

for second

we have

f(x)=x²-9x-36

let

y=x²-9x-36

when

x=0

y=-36

when

y=0

0=x²-9x-36

x²-12x+3x-36=0

x(x-12)+3(x-12)=0

(x-12)(x+3)=0

either

x=12

x=-3

So,

x and y intercepts are (-3,0)(12,0) and 0,-36)

7 0

2 years ago

The picture is there

aliina [53]

Answer:

Step-by-step explanation:

Because CB = 6 cm, we can find CD

Use Triangle CDB.

<BCD = <BCA - <ACD

<BCD = ?

<BCA = 90

<ACD = 60

<BCD = 90 - 60

<BCD = 30

Cos 30 = CD / CB

CD = Cos(30) * BC

CD = 5.196 cm

<A = 90 - ACD

<ACD = 60

<A = 90 - 60

<A = 30

Sin(<A) = CB / AB

AB = CB / sin(<A)

AB = 6 / 0.5

AB = 12

Area =1/2 CD * AB

Area = 1/2 * 5.196 * 12

Area = 31.18

4 0

2 years ago

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 12.1 fluid ounc

Ludmilka [50]

Answer:

$\mu = 12.1\\\sigma = 0.3$

a) Find the probability that the drink is less than 11.9 fluid ounces

We are supposed to find P(x<11.9)

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

Refer the z table

So, P(z<-0.67) = 0.2514

Hence the probability that the drink is less than 11.9 fluid ounces is 0.2514

b)Find the probability that the drink is between 11.6 and 11.9 fluid ounces

x= 11.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.6-12.1}{0.3}$

$z=-1.67$

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

We are supposed to find P(11.6<x<11.9)

So, P(11.6<x<11.9)= P(x<11.9)-P(x<11.6)

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

Refer the z table

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

P(-1.67<z<-0.67)= 0.2514-0.0475

P(-1.67<z<-0.67)= 0.2039

So, the probability that the drink is between 11.6 and 11.9 fluid ounces is 0.2039.

c)Find the probability that the drink is more than 12.6 fluid ounces

We are supposed to find P(x>12.6)

x= 12.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{12.6-12.1}{0.3}$

$z=1.67$

Refer the z table

P(x>12.6)=1-P(x<12.6)

P(z>1.67)=1-P(z<1.67)

P(z>1.67)=1-0.9525

P(z>1.67)=0.0475

Hence the probability that the drink is more than 12.6 fluid ounces is 0.0475

d) Is the drink containing more than 12.6 fluid ounces an unusual event?

Since p value is less than alpha (0.05)

Answer = No, because the probability that a drink containing more than 12.6 fluid ounces is less than 0.05, this event is not unusual.

3 0

3 years ago

For a certain candy, 20% of the pieces are yellow, 15% are red, 20% are blue, 20% are green, and the rest are brown. a

OLEGan [10]

Answer:

Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

Brown: $\frac{25}{100}$
Yellow or Blue: $\frac{20}{100} +\frac{20}{100} = \frac{40}{100}$ ....add the numerators

Not Green: $\frac{80}{100}$ .... since the sum of all the rest is 80%

Stiped: $\frac{25}{100}$ .... there are 0 striped candies.

Assuming the <u><em>ratios/percentages</em></u> of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.

3 Browns: $\frac{25*25*25}{100*100*100} = \frac{15,625}{1,000,000} = \frac{1.5625}{100}$

the 1st and 3rd are red while the middle is any. We multiply 15% * (total of all minus red which is 85%) * 15% like so.

$\frac{15*85*15}{100*100*100} = \frac{19,125}{1,000,000} = \frac{1.9125}{100}$

None are Yellow: multiply the percent of all minus yellow three times.

$\frac{80*80*80}{100*100*100} = \frac{512,000}{1,000,000} = \frac{51.2}{100}$

At least 1 green: multiply the percent of green by 100% twice, since the other two can by any

$\frac{20*100*100}{100*100*100} = \frac{200,000}{1,000,000} = \frac{20}{100}$

4 0

2 years ago