9^ ? 3^2 + 7^2
81 ? 9 + 49
81 > 58
c^ > a^2 + b^2
answer is obtuse triangle
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.
<span><span>1. </span>We
have 2 boxes weighs:
=> 9.4 lb and 62.6 lbs.
Now, let’s estimate the total weight of this 2 boxes using rounding and
compatible numbers.
For 9.4 lbs , the rounded and compatible number is 9 lbs
And for 62.6 lbs, the rounded and compatible number is 63 lbs
=> now, let’s try adding both numbers if we get a close answer
=> 9 +63
=> 72 = the estimated and rounded answer is 72.
Let’s check if we have close answer
=> 9.4 + 62.6
=> 72</span>
34/100
= 0.34
3 tenths, 4 hundreths
Answer:
find total volume to find what amount is 20%
hack, is put the 20% in alreadyy to find 20% volume
20%=1/5
so
vcone=1/3(hpir^2)
times thath by 1/5
vcone=1/15(hpir^2)
d/2=r
given
4.8=d
4.8/2=d/2=r=2.4
10=h
V=1/15(10pi2.4^2)
V=1/15(10pi5.76)
V=1/15(57.6pi)
V=3.84pi
V=12.0637 cubic inches=20% of vase
0.5 cubc inches per min
0.5 times t=12.0637
divide both sides by 0.5
t=24.1237
about 24 minutes
Step-by-step explanation:
