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3 years ago

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Daisy cut a square out of a sheet of graph paper. The square had an area of 16 square cm. She then trimmed 1 cm from each side o

f the square. What is the area of the snaler square?

1 answer:

Degger [83]3 years ago

7 0

Answer: 4 cm^2, it went from 4x4 to 2x2.. clipped on all sides...

Hope I helped!

Giving me brainliest is much appreciated! :).

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___ cosb =1/2 sin(a+b)+sin(a-b)?

vodomira [7]

Answer:

That would be sina.

Step-by-step explanation:

sin(a+b) = sinacosb + cosasinb

sin(a-b) = sinacosb - cosasinb

Adding we get sin(a+b) + sin(a-b) = 2sinaccosb

so sinacosb = 1/2sin(a+b) + sin(a-b)

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3 years ago

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Y^-1 dy +ye^(cosx) sinxdx=0

olga nikolaevna [1]

Please see the answer here

http://www.wolframalpha.com/input/?i=y%5E-1%20dy%20%2Bye%5E%28cosx%29%20sinxdx%3D0

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3 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago

a researcher found that a cigarette smoker smokes on average 31 cigarettes a day. she feels that this average is too high. she s

Ivenika [448]

At 0.01 significance level the hypothesis is rejected.

Null Hypothesis (H₀): The sample data occurs purely from chance.

Alternative Hypothesis (H₁): The sample data is influenced by some non-random cause.

If the p-value of the hypothesis test is less than some significance level (e.g. α = .01), then we can reject the null hypothesis and conclude that we have sufficient evidence to say that the alternative hypothesis is true.

Given that,

x = 28

s = 2.9

μ = 31

n = 8

α = 0.01

Then, the Null Hypothesis : H₀ = 31

The Alternate Hypothesis : H₁ < 31

The average number of cigeratte smoker per day is less than 31.

Test Statistics:

t = x -μ / s√n

= 28 - 31/ 2.9√8

= -3/ 8.41

= -0.356

Since the p-value of 0.0046 is less than the significance level of 0.01, the auditor rejects the null hypothesis.

Learn more about Null hypothesis:

brainly.com/question/28920252

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4 0

1 year ago

A measurement of the circumference of a disk has an uncertainty of 1 . 5 mm. How many measurements must be made so that the diam

Naya [18.7K]

Answer:

How many measurements must be made = 9

Step-by-step explanation:

The steps are as shown in the attachment.

3 0

3 years ago