Which phrase matches the algebraic expression below? 2(x − 7) + 10
1 answer:
<u>Answer:</u>
Two times the difference of a number x and 7 plus 10.
<u>Step-by-step explanation:</u>
We are to write a phrase which matches the following algebraic expression below:
We can see that there is a bracket with a coefficient outside it so we can express it as:
<em>'Two times the difference of a number x and 7'.</em>
While + 10 can be added to it to complete the phrase:
Two times the difference of a number x and 7 plus 10.
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I don't know how to do it except for with deritivies
so take the deritivieve and find where the deritivieve equals 0
that is where the sign changes
where the sign changes from (+) to (-), that is max
so
A.
max revenue
R'(x)=
find where numerator is 0
at x=6
to find change of sign, evaluate the denomenator at above and below 6 and see sign
R'(5)=(+)
R'(7)=(-)
at x=6, the sign changes from (+) to (-)
max is at x=6
sub 6 for x in the R(x) function
R(6)=9 (it's in thousands so $9000 is te max revenue)
B.
max profit
combine them
P(x)=R(x)-E(x)
take the deritive of P(x)
using sum rule
P'(x)=R'(x)-E'(x)
we already know what R'(x) is
E'(x)=
P'(x)=
find zeroes or what value of x make P'(x) equal to 0
use calculator or something or work it out to find x
at x=5.225
x is hundreds so times 100
522.5
about 523 items
A. $9000
B. 523 items
so take the deritivieve and find where the deritivieve equals 0
that is where the sign changes
where the sign changes from (+) to (-), that is max
so
A.
max revenue
R'(x)=
find where numerator is 0
at x=6
to find change of sign, evaluate the denomenator at above and below 6 and see sign
R'(5)=(+)
R'(7)=(-)
at x=6, the sign changes from (+) to (-)
max is at x=6
sub 6 for x in the R(x) function
R(6)=9 (it's in thousands so $9000 is te max revenue)
B.
max profit
combine them
P(x)=R(x)-E(x)
take the deritive of P(x)
using sum rule
P'(x)=R'(x)-E'(x)
we already know what R'(x) is
E'(x)=
P'(x)=
find zeroes or what value of x make P'(x) equal to 0
use calculator or something or work it out to find x
at x=5.225
x is hundreds so times 100
522.5
about 523 items
A. $9000
B. 523 items
Answer:
Yes. At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.
The random variable is the sample mean amount of mercury in the bass fish from the lakes of Florida.
The population parameter is the mean amount of mercury in the bass fish of Florida lakes.
The alternative hypothesis (Ha) states that the amount of mercury significantly differs from 1 mg/kg.
The null hypothesis (H0) states that the amount of mercury is not significantly different from 1 mg/kg.
Step-by-step explanation:
<em>The question is incomplete.</em>
<em>There is no data provided.</em>
<em>We will work with a sample mean of 0.95 mg/kg and sample standard deviation of 0.15 mg/kg to show the procedure.</em>
<em />
This is a hypothesis test for the population mean.
The claim is that the fish in all Florida lakes have different mercury than the allowable amount (1 mg of mercury per kg of fish).
Then, the null and alternative hypothesis are:
The significance level is assumed to be 0.05.
The sample has a size n=53.
The sample mean is M=0.95.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.15.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a two-tailed test, with 52 degrees of freedom and t=-2.427, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.019) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.