Answer:
To convert from degree to radian we multiply the angle by
π/180
To convert from radian to degree we multiply the angle by
180/π
5π/6,
Multiply by
180/π
=
5π/6 × 180/π = 5 × 30 = 150 degrees.
⇒5π/6 radian =150 degree.
Answer:
y = 4x + 14
Step-by-step explanation:
slope-intercept form: y = mx + b
Slope formula: 
To write the equation in y = mx + b form, we need to find the slope(m) and the y-intercept(b) of the equation.
To find the slope, take two points from the table(in this example I'll use points (0, 14) and (1, 18)) and input them into the slope formula:
Simplify:
18 - 14 = 4
1 - 0 = 1
The slope is 4.
To find the y-intercept, input the values of the slope and one point(in this example I'll use point (1, 18)) into the equation format and solve for b:
y = mx + b
18 = 4(1) + b
18 = 4 + b
14 = b
The y-intercept is 14.
Now that we know the slope and the y-intercept, we can write the equation:
y = 4x + 14
I will assume you are using compound interest.
<span>let the amount invested be x </span>
<span>x(1.0575)^25 = 85000 </span>
<span>x = 85000/1.0575^25 = $21,009.20</span>
Helen runs the longest distance in 4 days.
Helen (In 4 days): 2.75 x 4 = 11
Bill (In 4 days): 10
11 (Helen) > 10 (Bill), so therefore Helen runs the longest distance in 4 days.
Answer:
A - 0%
B- 50%
C- 50%
D- 100%
Step-by-step explanation:
Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.
Performing the cross of Aa x AA, we can see that:
a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.
b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.
c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.
d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.
