Answer:
Step-by-step explanation:
y = -4x + 5.....slope here is -4. A parallel line will have the same slope.
y = mx + b
slope(m) = -4
(-6,-2)....x = -6 and y = -2
now we sub and find b, the y intercept
-2 = -4(-6) + b
-2 = 24 + b
-2 - 24 = b
-26 = b
so ur equation is : y = -4x - 26 <====
-8(1 + 8n) - 8(6 - 4n) = -24
-8 - 64n - 48 + 32n = -24
-32n - 56 = -24
-32n = 32
n = -1
Answer:
X =8
EF=12
FG=17
Step-by-step explanation:
Answer:
The probability of using one or the other is 36%
Step-by-step explanation:
For solving this problem it is easy if we see it in a ven diagram, for this first we are going to name the initial conditions with some variables:
Probability of passing Professor Jones math class = 64% =0,64
P(J) = 0.64
Probabiliry of passing Professor Smith's physics class = 32% =0.32
P(S) = 0.32
Probability of passing both is = 30% = 0.30
P(JnS) = 0.30 (Is is an intersection so it is in the middle of the ven diagram
We need to know which is the probability of pasing one or the other for this we need to take out the probability of passing both for this we have to add the probability of passing Professor Jones math class with the probabiliry of passing Professor Smith's physics class and substract the probability of passing both for each one:
P(JuS) = (P(J) - P(JnS)) + (P(S) - P(JnS)) = (0.64 - 0.30) + (0.32 - 0.30) = 0.34 + 0.02 = 0.36 = 36%
If you check the ven diagram you can see that if we add all what is in red we will have the probability of passing Professor Jones math class and if we add all what is in blue we wiill have the probability of passing Professor Smith's physics class, and if we add just what is in each corner we will get the same value that is the probabilty of passsing one or the other.
Answer:
z= 3.63
z for significance level = 0.05 is ± 1.645
Step-by-step explanation:
Here p = 42% = 0.42
n= 500
We formulate our null and alternative hypotheses as
H0: p= 0.42 against Ha : p> 0.42 One tailed test
From this we can find q which is equal to 1-p= 1-0.42 = 0.58
Taking p`= 0.5
Now using the z test
z= p`- p/ √p(1-p)/n
Putting the values
z= 0.5- 0.42/ √0.42*0.58/500
z= 0.5- 0.42/ 0.0220
z= 3.63
For one tailed test the value of z for significance level = 0.05 is ± 1.645
Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.