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Agata [3.3K]
3 years ago
11

Write 3x+4y=12 in slope intercept form

Mathematics
2 answers:
cestrela7 [59]3 years ago
5 0
Ok, let’s take it step by step.
In order to write an equation in slope intercept form, solve for y.
3x+4y=12
First, subtract 3x from both sides.
4y=-3x+12
Now, divide both sides by 4.
y=-3/4x+3.

So, 3x+4y=12 written in slope intercept form is y=-3/4x+3.
lapo4ka [179]3 years ago
3 0

Answer:

y= -3/4x+12

Step-by-step explanation:

subtract 3x = 4y= -3x + 12

divide by 4

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600 k (Kilograms to grams is:
 600000 g (grams)
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3 years ago
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A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading
valkas [14]

Answer:

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

p_v =2*P(t_{11}  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}

The results obtained are:

\bar X=98.375 represent the sample mean  

s=0.6.109 represent the sample standard deviation  

n=12 sample size  

\mu_o =100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

df=n-1=12-1=11

Since is a two sided test the p value would given by:  

p_v =2*P(t_{11}  

5) Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

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3 years ago
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2 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

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