The bakery start with 2954 cherry
<em><u>Solution:</u></em>
Given that,
A bakery made 26 cherry pies, using 115 cherries for each pie
So they used 115 cherry for each pie
26 cherry pies, 115 cherry per pie
Therefore,
Number of cherries used = 26 x 115 = 2990 cherries used
They threw away 36 cherries that were bad
36 cherry were bad
2990 - 36 = 2954
Thus they start with 2954 cherries
9514 1404 393
Answer:
13064 feet
Step-by-step explanation:
We can rewrite the formula so that x is in feet. Equating the new formula to 8.743 PSI, we have ...
8.743 = 14.7e^(-0.21x/5280)
Dividing by 14.7 and taking natural logs, we have ...
ln(8.743/14.7) = -0.21x/5280
Multiplying by the inverse of the coefficient of x gives ...
x = 5280·ln(8.743/14.7)/-0.21 ≈ 13064.0806
The peak of the mountain is about 13,064 feet high.
Step-by-step explanation:
<h3>Let the literate and illiterate people of Rashkundu Village be 4x and x respectively then </h3><h3>4x + x = 6550</h3><h3>5x = 6550</h3><h3>x = 6550 / 5</h3><h3>x = 1310</h3><h3 /><h3>Therefore </h3><h3>no. of literate people = 4 * 1310 = 5240</h3><h3 /><h3>no. of illiterate people = 1310 </h3><h3 /><h3>Hope it will help :)</h3>
The answer is A because you turn the fraction in a decimal then divide by 5
Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
75% Confidence Interval can be calculated using M±ME where
- M is the sample mean weight of the wild mountain lions (
)
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME=
where
- t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
- s is the standard deviation of the sample(31.4)
Thus, ME=
≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5