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ivann1987 [24]
3 years ago
15

What percent if 6 is 1.2

Mathematics
2 answers:
Nataliya [291]3 years ago
6 0
What percentage of 6 is 1.2

\dfrac{1.2}{6}  \times 100 = 20%

Answer: 1.2 is <u>20%</u> of 6
natulia [17]3 years ago
3 0
Answer:  "<u /> <u>20 % </u>" of 6  is 1.2 " . 
______________________________________________________
Explanation:
______________________________________________________
" What percent OF "6" is "1.2" ? "
______________________________________________________
{ n/100 } * 6 = 1.2 ;   Solve for "n" ; 

Divide each side of the equation by "6" ; 

to get: 

n/100 = 1.2 / 6 ; 

→ n / 100 = 0.2 ; 

Multiply each side of the equation by "100" ; 
    to isolate "n" on one side of the equation; & to solve for "n" ; 

→ n = (0.2) * 100 ; 

→ n = "20" . 

Answer:  "20 % of 6 is 12 " . 
_______________________________________
Alternate method:

(n/100) * 6 = 1.2 ; 

(n/100) * (6/1) = 1.2 ; 

6n/ 100 = 1.2 ; 

6n = 1.2 * 100  ;

6n = 120 ; 

Divide each side of the equation by "6" ; 
to isolate "n" on one side of the equation; & to solve for "n" ; 

6n / 6 = 120 / 6 ; 

n = "20" 
_________________________
→  "20 %" of 6 is 1.2 " . 
_________________________
Alternate method:
__________________________
n/100   * 6 = 1.2 ; 

→ 6n / 100 = 1.2 ; 

→ 3n / 50 = 1.2 ; 

→ 3n = 1.2 * 50 ; 

→ 3n =  60.0

→ 3n = 60 ; 

Divide each side of the equation by "3" ; 
to isolate "n" on one side of the equation; & to solve for "n" ; 

→  3n/ 3 = 60 / 3 ; 

→ n = "20" ; 

Answer:  " 20 % " of 6 is 1.2 " . 
__________________________________________________
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Answer:

15√15−15√5+15√3−15

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Exact Form: 15√15−15√5+15√3−15

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2 years ago
Find the P-value for the hypothesis test with a standardized test statistic z. Decide whether toreject the null hypothesis for t
iren2701 [21]

Answer:

1.) We don't reject the null

2.) We reject the Null

3.) We do not reject the null

Step-by-step explanation:

Obtaining p values using test statistic:

Given that ; we have a standardized test statistic and α - values ;

Let's define the decision region :

If Pvalue < α ; Reject H0 ; otherwise, fail to reject H0

A.)

Left tail , Z = - 1.32 ; α = 0.10

We can use the Pvalue calculator from Z score :

Pvalue = 0.934

Pvalue > α ; Hence, we fail to reject the null, H0

B.)

Right-tailed test, z = 2.46, α = 0.01

Pvalue from Zscore calculator ;

Pvalue = 0.0069

Pvalue < α ; Hence, we reject the null, H0

C.)

Two-tailed test, z = -1.68, α = 0.05

Pvalue from Zscore calculator ;

Pvalue = 0.093

Pvalue > α ; Hence, we fail to reject the null, H0

6 0
3 years ago
A scientist testing the effects of a chemical on apple yield (apples/acre) sprays an orchard with the chemical. A second orchard
JulsSmile [24]

Following are the dependent variables:

<em>1. The amount of water that each orchard receives.</em>

<em>2. The species of trees in the orchard.</em>

Reason:

The exercise scientist is looking for the effects of a chemical between an apple crop to which it is administered and another to which it is not, 4 options are presented, of which it is essential to count as a variable the amount of water each Orchard and tree species in the orchard, since they can generate alterations in the results, the other two variables of the exercise such as number of apples and size of the orchards are not significant and their variations do not affect the scientist's objective.

Learn more about Dependent Variable on:

brainly.com/question/1670595

#SPJ4

3 0
2 years ago
An urn contains 2 red marbles and 3 blue marbles. 1. One person takes two marbles at random from the urn and does not replace th
Ghella [55]

Answer:

A) The best way to picture this problem is with a probability tree, with two steps.

The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.

The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.

If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.

If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.

B) So a person can have a red marble and a blue marble in two ways:

1) Picking the red first and the blue last

2) Picking the blue first and the red last

C) P(R&B) = 3/5 = 60%

Step-by-step explanation:

C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5

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