ANSWER:
For the first picture, it is the first answer. This is because of the Exterior Angle Theorem.
For the second picture, use the Exterior Angle Theorem again, but this time, solve for x.
The m∡B = 39°
Answer:
200
Step-by-step explanation:
2×2×2=8
5×5=25
8×25=200
Answer:
c. 0.1151
Step-by-step explanation:
d) the probability that, on average, fish are larger than 86 centimeters in length.The area under part of a normal probability curve is directly proportional to probability and the value is calculated as
z = (x₁−x) /σ
where z = propability of normal curve
x₁ = variate mean = 86cm
x = mean of 80cm
σ = standard deviation = 5cm
applying the formula,
z= (86-80)/5
z = 6/5 =1.2
Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 1.2 corresponds to an area of 0.3849 between the mean value. but, because the standard curve has 0.5, then will minus 0.3849 from 0.5= 0.5 - 0.3849 = 0.1151
Thus the probability of a fish are larger than 86 centimeters in length is 0.1151
The annual interest rate is 3.5%.
Solution:
Given Interest (I) = $26.25
Principal (P) = $500
Time (t) = 18 months
Rate of interest (r) = ?
Time must be in years to find the rate per annum.
1 year = 12 months
Divide the time by 12.
Time (t) = 
years
Now, find the rate of interest using simple interest formula.
<u>Simple interest formula:</u>
⇒ r = 3.5%
Hence the annual interest rate is 3.5%.
Let us model this problem with a polynomial function.
Let x = day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.
Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅
Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
| 1 1 1 1 1 | | a₁ | | 42 |
| 2⁴ 2³ 2² 2¹ 2⁰ | | a₂ | | 26 |
| 3⁴ 3³ 3² 3¹ 3⁰ | | a₃ | = | 61 |
| 4⁴ 4³ 4² 4¹ 4⁰ | | a₄ | | 65 |
| 5⁴ 5³ 5² 5¹ 5⁰ | | a₅ | | 56 |
When this matrix equation is solved in the calculator, we obtain
a₁ = 4.1667
a₂ = -55.3333
a₃ = 253.3333
a₄ = -451.1667
a₅ = 291.0000
Test the solution.
y(1) = 42
y(2) = 26
y(3) = 61
y(4) = 65
y(5) = 56
The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.
Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631
