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.
Draw the picture and label the
width = w
The length of the monitor is six times the quantity of five less than half its width:
length = 6(w/2-5)
length = 3w-30
Area = (length)*(width)
384=(3w-30)*(w)
384=3w^2-30w
Answer:
the dimensions of the monitor in terms of its width is:
384=3w^2-30w
Answer: Adiya’s method is not correct. To form a perfect square trinomial, the constant must be isolated on one side of the equation. Also, the coefficient of the term with an exponent of 1 on the variable is used to find the constant in the perfect square trinomial. Adiya should first get the 20x term on the same side of the equation as x2. Then she would divide 20 by 2, square it, and add 100 to both sides.
Lois is correct. This is because multiplication is repeated addition. It means that when your addends are just the same number but added several times, the sum is just equivalent to the product of that number times the number of times it was added. For example, 4+4+4 = 4×3 = 12.
Answer: you need to have a line graph
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