Answer:
Here we have given two catogaries as degree holder and non degree holder.
So here we have to test the hypothesis that,
H0 : p1 = p2 Vs H1 : p1 not= p2
where p1 is population proportion of degree holder.
p2 is population proportion of non degree holder.
Assume alpha = level of significance = 5% = 0.05
The test is two tailed.
Here test statistic follows standard normal distribution.
The test statistic is,
Z = (p1^ - p2^) / SE
where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2]
p1^ = x1/n1
p2^ = x2/n2
p^ = (x1+x2) / (n1+n2)
This we can done in TI_83 calculator.
steps :
STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER
Test statistic Z = 1.60
P-value = 0.1090
P-value > alpha
Fail to reject H0 or accept H0 at 5% level of significance.
Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.
Answer:
The number of strawberry baskets sold was 44
Step-by-step explanation:
Let
x ----> the number of strawberry baskets sold
y ---> the number of raspberries baskets sold
we know that
In one morning, Katya sells 96 baskets for $644
so
----> equation A
---> equation B
Solve the system by graphing
Remember that the solution is the intersection point both graphs
using a graphing tool
The solution is the point (44.52)
see the attached figure
therefore
The number of strawberry baskets sold was 44 and the number of raspberries baskets sold was 52
First, you divide 58 bars by 4. The answer will be 14 boxes with 2 bars remaining in the box.
The sum of a number b and 3 is greater than 8 and less than
12
<span>=> b + 3 < 8
=> b + 3 > 12
</span>
<span>Now,
let’s combine this into an expression:
=> 8 < b+3 < 12
=> (8 - 3) < b < (12 - 3)
=> 5 < b < 9
Thus, the answer in the given situation is 9 because 9 is greater than 8 but
less than 12.
</span>
Answer:
D
Step-by-step explanation:
360h+6,400 >8,020, with a solution is h> 4.
