Answer:
9 - 7x - x²
Step-by-step explanation:
-4x+2x2-x²+5-3x = -4x-3x+2x2+5-x²
I arranged the like terms together so it is easier to do the calculations.
-4x-3x+2x2+5-x² = -7x+4+5-x²
Simplify.
-7x+4+5-x² = 9-7x-x²
The bacteria grow from 1,000 bacteria into 12,500 bacteria in 12 hours. The exponent function would be
:an= ao(1+r)^t12.500= 1,000(1+r)^121,000(12.5)= 1,000(1+r)^1212.5= (1+r)^121+r= 1.23427r= 0.23Number of bacteria after 20hr would be:an= ao(1+r)^tan= 1,000(1+0.23)^20an= 1,000(62.821 )an= 62,821
So attached is a picture of the triangle you are talking about and listed under are the choices:
A.) Cos Z=b/c
B.) Sin X=c/b
C.) Tan X=b/a
<span>D.) Tan Z=a/b
</span>
The answer would then be
B. SinX = c/b.
Just remember SOH CAH TOA:
Sinθ= Opposite Cosθ = Adjacent Tanθ= Opposite
Hypotenuse Hypotenuse Adjacent
Using the triangle in the scenario, you just need to identify which side is which.
Given m∠ZAdjacent = b
Opposite = a
Hypotenuse = c
SinZ= a CosZ = b TanZ= a
c c b
Given m∠X:
Adjacent = b
Opposite = a
Hypotenuse = c
SinX= <u> b </u> CosX =<span><u> a </u></span> TanX=<span><u> b </u></span>
c c a
So the answer is B.
Attached is a picture of how I assigned the sides depending on the angle used.
Answer:
1. D. 20, 30, and 50
2. A. 86
3. B. 94
Step-by-step explanation:
1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.
The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.
Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.
Thus:
Q1 = (85 + 87)/2 = 86
Q3 = (93 + 95)/2 = 94
IQR = Q3 - Q1 = 94 - 86
IQR = 8
Outliers in the data set are data values below the lower limit or above the upper limit.
Let's find the lower and upper limit.
Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74
The data values below the lower limit (74) are 20, 30, and 50
Let's see if we have any data value above the upper limit.
Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106
No data value is above 106.
Therefore, the only outliers of the data set are:
D. 20, 30, and 50
2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.
Thus:
Q1 = (85 + 87)/2 = 86
3. Q3 = (93 + 95)/2 = 94