answer is Lcm which is probaly c
Answer:
13
Step-by-step explanation:
4 + 3(x + 5)=5 x - 7
3x + 19=5x - 7
3x + 19 - 19=5x - 7 - 19
3x =5x - 26
3x - 5x=5x - 26 - 5x
-2= -26
-2x/-2 = -26/-2
x= 13
It might be wrong. I may have made a mistake with typing sorry.
Hope this helps a bit.
Hello!
The formula for finding area is l × w
If the area is 19 ½ inches and it was 3 ¼ inches wide, then the length will be 19 ½ ÷ 3 ¼ (Area ÷ Width)
![19 \frac{1}{2} \div 3 \frac{1}{4}](https://tex.z-dn.net/?f=19%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Cdiv%203%20%20%5Cfrac%7B1%7D%7B4%7D%20%20)
Change both to improper fraction
![19 \frac{1}{2} = \frac{39}{2}](https://tex.z-dn.net/?f=19%20%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%20%20%5Cfrac%7B39%7D%7B2%7D%20)
![3 \frac{1}{4} = \frac{13}{4}](https://tex.z-dn.net/?f=3%20%20%5Cfrac%7B1%7D%7B4%7D%20%3D%20%20%5Cfrac%7B13%7D%7B4%7D%20%20)
![\frac{39}{2} \div \frac{13}{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B39%7D%7B2%7D%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B4%7D%20%20)
Keep 39/2
Change ÷ to ×
Flip 13/4 = 4/13
![\frac{39}{2} \times \frac{4}{13} = \frac{156}{26} = 6](https://tex.z-dn.net/?f=%20%5Cfrac%7B39%7D%7B2%7D%20%5Ctimes%20%20%5Cfrac%7B4%7D%7B13%7D%20%3D%20%20%5Cfrac%7B156%7D%7B26%7D%20%3D%206)
Hello :D
To work this out you need to round the numbers to a whole number.
19.9 = 20
5.8 = 6
20 x 6 ≈ 120
I hope this helps :)
Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:
![z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ z-score =2.326](https://tex.z-dn.net/?f=z-score%3D%20%5Cfrac%7B90-83.6%7D%7B%20%5Cfrac%7B8.7%7D%7B%20%5Csqrt%7B10%7D%20%7D%20%7D%20%5C%5C%20%20%5C%5C%20%0Az-score%20%3D2.326%20)
So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.