PEMDAS tells us to use multiplication and division before addition and subtraction.
<u>-5(8) </u>÷ 10 - <u>14 ÷ 7</u> - <u>(-2)(3)</u>
<u> -40 ÷ 10</u> - 2 - -6
<u>-4 - 2 </u> + 6 <em>two negatives make a positive so - - 6 = + 6</em>
<u> -6 +6</u>
0
Answer: 0
Answer:
2nd choice OC.y=-2x+3
Step-by-step explanation:
The equation for the slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept. With these two numbers both given to us, we can just plug them in and get y=-2x+3.
Hope this helps you
#1) 5/20
#2) 7/300
#3) 3/50
#4) 5/20
Explanation
For #1:
There are 15 even numbers out of 30. Since it is replaced before drawing the second ball, there will be 15 odd numbers out of 30. This gives us
15/30(15/30) = 225/900 = 5/20.
For #2:
There is 1 7 out of 30; then there are 14 numbers greater than 16 out of 30:
1/30(14/30) = 14/900 = 7/300
For #3:
There are 6 multiples of 5 out of 30; then there are 9 prime numbers out of 30:
6/30(9/30) = 54/900 = 3/50
For #4:
There are 15 even numbers out of 30; then there are still 15 even numbers out of 30:
15/30(15/30) = 225/900 = 5/20
Answer:
Answer:
Since the calculated value of t= -1.340 does not fall in the critical region , so we accept H0 and may conclude that the data do not provide sufficient evidence to indicate hat there is difference in mean carbohydrate content between "meals with potatoes" and "meals with no potatoes".
Step-by-step explanation:
Potatoes : No Potatoes : Difference Difference (d)²
(Potatoes- No Potatoes)
29 41 -12 144
25 41 -16 256
17 37 -20 400
36 29 -7 49
41 30 11 121
25 38 -13 169
32 39 -7 49
29 10 19 361
38 29 9 81
34 55 -21 441
24 29 -5 25
27 27 0 0
<u>29 31 -2 4 </u>
<u> ∑ -64 2100 </u>
- We state our null and alternative hypotheses as
H0 : μd= 0 and Ha: μd≠0
2. The significance level alpha is set at α = 0.01
3. The test statistic under H0 is
t= d`/sd/√n
which has t distribution with n-1 degrees of freedom.
4. The critical region is t > t (0.005,12) = 3.055
5. Computations
d`= ∑d/n = -64/ 13= -4.923
sd²= ∑(di-d`)²/ n-1 = 1/n01 [ ∑di² - (∑di)²/n]
= 1/12 [2100- ( -4.923)] = 175.410
sd= √175.410 = 13.244
t = d`/sd/√n= - 4.923/13.244/√13
t= - 4.923/3.67344
t= -1.340
6. Conclusion :
Since the calculated value of t= -1.340 does not fall in the critical region , so we accept H0 and may conclude that the data do not provide sufficient evidence to indicate hat there is difference in mean carbohydrate content between "meals with potatoes" and "meals with no potatoes".
