Isabel can drive 505.85 miles with a gas tank that holds 15.1 gallons of gas
<em><u>Solution:</u></em>
Given that, Isabel’s car gets 33.5 miles per gallon
Her gas tank holds 15.1 gallons of gas
To find: Number of miles can Isabel drive on a full tank of gas
Let "x" be the miles drive with 15.1 gallons of gas
From given,
1 gallon = 33.5 miles
15.1 gallon = x miles
This forms a proportion and we can solve the sum by cross multiplying
Thus Isabel can drive 505.85 miles on a full tank of gas
Answer:
Option A
The p-value is less than the significance level of 0.05 chosen and so we reject the null hypothesis H0 and can conclude that the proportion of the subjects who have the necessary qualities is less than 0.2.
Step-by-step explanation:
Normally, in hypothesis testing, the level of statistical significance is often expressed as the so-called p-value. We use p-values to make conclusions in significance testing. More specifically, we compare the p-value to a significance level "α" to make conclusions about our hypotheses.
If the p-value is lower than the significance level we chose, then we reject the null hypotheses H0 in favor of the alternative hypothesis Ha. However, if the p-value is greater than or equal to the significance level, then we fail to reject the null hypothesis H0
though this doesn't mean we accept H0 automatically.
Now, applying this to our question;
The p-value is 0.023 while the significance level is 0.05.
Thus,p-value is less than the significance level of 0.05 chosen and so we reject the null hypothesis H0 and can conclude that the proportion of the subjects who have the necessary qualities is less than 0.2.
The only option that is correct is option A.
Answer:
Step-by-step explanation:
A fraction can be found be taking the current amount over the total amount:
Since Darlene has only typed 23 pages out of 59, her fraction is 23 over 59. Fractions must always be in simplest form, which means that both the numerator and denominator can no longer be divided by the same factor. Since the numbers 23 and 59 are both prime and have no other factors other than 1 and themselves, the fraction 23/59 is in simplest form.
There are many polynomials that fit the bill,
f(x)=a(x-r1)(x-r2)(x-r3)(x-r4) where a is any real number not equal to zero.
A simple one is when a=1.
where r1,r2,r3,r4 are the roots of the 4th degree polynomial.
Also note that for a polynomial with *real* coefficients, complex roots *always* come in conjugages, i.e. in the form a±bi [±=+/-]
So a polynomial would be:
f(x)=(x-(-4-5i))(x-(-4+5i))(x--2)(x--2)
or, simplifying
f(x)=(x+4+5i)(x+4-5i)(x+2)^2
=x^4+12x^3+77x^2+196x+164 [if you decide to expand]
Answer:
a) 30 kangaroos in 2030
b) decreasing 8% per year
c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo
Step-by-step explanation:
We assume your equation is supposed to be ...
P(t) = 76(0.92^t)
__
a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30
In the year 2030, the population of kangaroos in the province is modeled to be 30.
__
b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.
The population is decreasing by 8% each year.
__
c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.
P(100) = 75(0.92^100) = 76(0.0002392)
P(100) ≈ 0.0182, about 1/55th of a kangaroo
