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allsm [11]
3 years ago
5

1. A jeepney driver claims that his average monthly income is Php 3000.00 with a standard deviation of Php 300.00. A sample of 3

0 jeepney drivers were surveyed and found that their average monthly income is Php 3500.00 with a standard deviation of Php 350.00. Test the hypothesis at 1% level of significance?​
Mathematics
1 answer:
nydimaria [60]3 years ago
7 0

Answer:

Kindly check explanation

Step-by-step explanation:

The hypothesis :

H0 : μ = 3000

H0 : μ ≠ 3000

The test statistic :

(xbar - μ) ÷ (s/√(n))

xbar = 3500

μ = 3000

σ = 300

n = 30

(3500 - 3000) ÷ (350/√(30))

Test statistic = 7.824

Df = 30 - 1 = 29

Tcritical at 0.01 = 2.462

Test statistic > critical value ; we reject H0 ; and concluded that there is significant evidence that

μ ≠ 3000

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116 if you add 72 plus 43
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Solve the problem. The half-life of plutonium-234 is 9 hours. If 50 milligrams is present now, how much will be present in 6 day
MaRussiya [10]

Answer:

  4)  0.001

Step-by-step explanation:

The amount (a) present in h hours is ...

  a(h) = (initial value)×(1/2)^(h/(half-life))

6 days is 6×24 hours, so the amount remaining is ...

  a(6×24) = 50×(1/2)^(6×24/9) = 50/2^16 ≈ 0.001 . . . . milligrams

3 0
3 years ago
Question 7 of 10
Ganezh [65]

Answer:

  • Option B

Step-by-step explanation:

Given Equation :

\qquad \sf \dashrightarrow \: 3(4x+3) = 2x - 5(3 - x) + 2

Using distribute property:

\qquad \sf \dashrightarrow \: 12x + 9 = 2x - 15 + 5x + 2

Adding the like terms we get :

\qquad \sf \dashrightarrow \: 12x + 9 = 2x  + 5x  - 15 + 2

\qquad \sf \dashrightarrow \: 12x + 9 = 7x  - 13

Transposing the variables on the right side and constant terms on the left side :

\qquad \sf \dashrightarrow \: 12x  - 7x =   - 13 - 9

\qquad \sf \dashrightarrow \: 5x =   - 22

Dividing both sides by 5 :

\qquad \sf \dashrightarrow \:  \dfrac{5x}{5}  =  \dfrac{ - 22}{5}

\qquad \bf \dashrightarrow \:  x =  \dfrac{ - 22}{5}

3 0
2 years ago
−4x+7y+5=0<br> x−3y=−5<br> ​ <br><br> How many solutions does the system have?
xeze [42]

   

Solution:  

Using Substitution Method:

-4x+7y=-5   (Equation 1)

x-3y=-5       (Equation 2)

get the value of x from Equation 2

x=3y-5     (Equation 3)  

Put the value of x from Equation 3 in Equation 1

-4(3y-5)+7y=-5

-4(3y)+20+7y=-5

-12y+7y=-5-20

-5y=-25

Negative sign on both sides cancels each other

y=25/5

y=5

Putting value of y in equation 3

x=3(5)-5

x=15-5

x=10

Therefore,   [x,y]=[10,5]

Using Elimination Method

-4x+7y=-5   (Equation 1)

x-3y=-5       (Equation 2)

Multiply equation 2 with -4 in order to eliminate the x term

-4(x-3y)=-5*4

-4x+12y=20     (Equation 3)

Adding Equation 1 and 3

-4x+7y=-5    

-4x+12y=20    

+    -     = -   (Change Of Sign with x and y terms)

-----------------

0x-5y = -25

-5y=-25

y=5  

Substituting y’s value is Equation 1

-4x+7(5)=-5

-4x+35=-5

-4x=-40

Cancellation of negative sign on both sides

x=40/4

x=10

[x,y]=[10,5]

3 0
3 years ago
Can i get help with this plss
damaskus [11]
49.32 multiply 0.09 x 548 to get the answe
3 0
3 years ago
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