Answer:
The symphonic choir can record a maximum of 14 songs.
Step-by-step explanation:
We know that the album cannot be more than 72 minutes long, and the jazz choir records a total of 
song minutes.
If we call 
the number of songs the symphonic choir records , and each song is 3.5 minute long, then the song minutes for the symphonic choir are 
; therefore, we have the inequality
<em> (this says the song minutes for jazz choir plus song minutes for symphonic choice cannot exceed 72 minutes )</em>
We solve this inequality by subtracting 21 from both sides and then dividing by 3.5:
The maximum integer value can take is 14; therefore, the maximum number of songs the symphonic choir can record is 14 songs.
Cost per ounce means you take the cost and divide it by the cost:
<span>3.36 / (21/2) </span>
<span>division of fractions changes to the multiplication of the reciprocal: </span>
<span>3.36 * (2 / 21) </span>
<span>3.36 and has 21 as a factor, so let's cancel it: </span>
<span>0.16 * 2 </span>
<span>$0.32 per ounce</span>
1/4
2/8
4/16
3/12
u can convert fractions into percents by multiplying the first number by the second. hope this helped:))
Answer:
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Step-by-step explanation:
The sample proportion is p2= 7/27= 0.259
and q2= 0.74
The sample size = n= 27
The population proportion = p1= 0.4
q1= 0.6
We formulate the null and alternate hypotheses that the new program is effective
H0: p2> p1 vs Ha: p2 ≤ p1
The test statistic is
z= p2- p1/√ p1q1/n
z= 0.259-0.4/ √0.4*0.6/27
z= -0.141/0.09428
z= -1.496
The significance level ∝ is 0.05
The critical region for one tailed test is z ≤ ± 1.645
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
What’s the full question?
