Answer:
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle? Here the number of animals and the number of days are in inverse proportion. Hence the food will last 4 days.
Answer:
1. -5
2. 0
Step-by-step explanation:
<em>1.</em> Distribute the Negative Sign:
=a−2a+−1(3a−(4a−5))
=a+−2a+−1(3a)+−1(−4a)+(−1)(5)
=a+−2a+−3a+4a+−5
Combine Like Terms:
=a+−2a+−3a+4a+−5
=(a+−2a+−3a+4a)+(−5)
Answer:
=−5
<em>2. </em>Distribute:
=(13)(5)+(13)(−t)+−5+t+(−12)(5)+(−12)(−t)
=65+−13t+−5+t+−60+12t
Combine Like Terms:
=65+−13t+−5+t+−60+12t
=(−13t+t+12t)+(65+−5+−60)
Answer:
=0
868 is about 870, and 28 is about 30, so your expression would be 870 divided by 30 which would be 29 and the actual answer is 31, so your estimate would be close to your actual answer.
Let x be the initial width, then the initial length is x+20.
<span>If Elise decreases this length by 8 feet, she will get the new length x+20-8=x+12.
</span>
If Elise increases the width by 10 feet, <span>she will get the new width x+10.
</span><span>
</span><span>The new perimeter will be x+12+x+12+x+10+x+10=4x+44=172, 4x=172-44, 4x=128, x=128÷4, x=32.
</span><span>
</span><span>The initial width is 32 ft and the initial length is 32+20=52 ft.
</span><span>
</span><span>The new width will be 32+10=42 ft, the new length will be 32+12=44 ft.
</span>
Answer:
Step-by-step explanation:
Given that:
Population Mean = 7.1
sample size = 24
Sample mean = 7.3
Standard deviation = 1.0
Level of significance = 0.025
The null hypothesis:
The alternative hypothesis:
This test is right-tailed.
Rejection region: at ∝ = 0.025 and df of 23, the critical value of the right-tailed test 
The test statistics can be computed as:
t = 0.980
Decision rule:
Since the calculated value of t is lesser than, i.e t = 0.980 < , then we do not reject the null hypothesis.
Conclusion:
We conclude that there is insufficient evidence to claim that the population mean is greater than 7.1 at 0.025 level of significance.
