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3 years ago

What’s the surface area to 1,2,3 and 4

Mathematics

1 answer:

topjm [15]3 years ago

4 0

Wdym? Can you insert a photo or something.. I’m confuse uh..

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Show how you got the answer

rewona [7]

Subtract the 45 minutes they were eating from 1:45pm since that’s when they arrived. You now have 1:00pm, you now just need to find out how long it is from 10:20 am to 1:00pm. HINT: it’s 2h 40m

5 0

3 years ago

2 Question

Margarita [4]

Answer:

Net Change in Geno's Field Position = 22 yards

Step-by-step explanation:

Assuming Geno initial position before change is 0

Geno Gained 4 yards three times = 4*3 = 12yards

Geno lose 1 yard twice = 1*(-2) = -2yards

Geno Gained 6 yards twice = 6*2 = 12yards

Net Change in Geno's Field Position = 12 + (-2) + 12 yards

= 22 yards

4 0

3 years ago

A fishing boat travels 36 miles with the current of two hours. It travels 42 miles against the current in three hours

nikitadnepr [17]

Answer:

42 miles in 3 hours it is the answer

7 0

3 years ago

The diagram shows the blueprint of a vegetable garden. All dimensions are given in yards.

Delicious77 [7]

Answer:

Step-by-step explanation:

i saw another brainly and he seemed to be correct and i got it right :)!

6 0

3 years ago

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Da

alexgriva [62]

Answer:

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

The question is incomplete:

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.

We start calculating the sample mean M and standard deviation s:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(18+12+20+17+14+15+13+11+21+17)\\\\\\M=\dfrac{158}{10}\\\\\\M=15.8\\\\\\$

$s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{\dfrac{101.6}{9}}\\\\\\s=\sqrt{11.3}=3.4\\\\\\$

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{10}}=\dfrac{3.4}{3.162}=1.075$

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=3.25 \cdot 1.075=3.49$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3$

The 99% confidence interval for the mean germination time is (12.3, 19.3).

8 0

3 years ago