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kati45 [8]
2 years ago
8

Please help.

Mathematics
2 answers:
ipn [44]2 years ago
8 0
2£ sorry if it's wronge
Archy [21]2 years ago
7 0

Answer:

2π

Step-by-step explanation:

When a circle's radius is 1—called a unit circle—its circumference is 2π.

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Expand the following expressions by multiplying out the brackets <br> -4(4f+3)
melisa1 [442]

Answer:

- 16f - 12

Step-by-step explanation:

- 4(4f + 3) ← multiply each term in the parenthesis by - 4

= - 16f - 12

6 0
2 years ago
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A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance tra
scZoUnD [109]

Answer:

4.038

Step-by-step explanation:

The simple regression model is:

Time = \hat\beta _{0} + \hat\beta_{1}Distance = -7.126 + 0.0214Distance, then \hat\beta_{1} = 0.0214.

The statistic calculated for the hypothesis test of statistical significance, for the slope, in a simple regression model is given by:

t_{c} = \frac{\hat\beta_{1}-\beta_{1_{0}}}{SE}

with \hat\beta_{1}  the slope estimator, \beta_{1_{0}}  the value of slope in the null hypothesis and SE the standard error of the slope. Thus,

t_{c} = \frac{0.0214-0}{0.0053} = 4.038

5 0
3 years ago
Jasmine's dessert takes 3½ minutes to cook in the microwave. <br><br> How many seconds is this?
Artemon [7]
210 seconds

1 minute = 60 seconds
4 0
3 years ago
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For a population with an unknown distribution, the form of the sampling distribution of the sample mean is _____.a. exactly norm
maxonik [38]

Answer:

Approximately normal for large sample sizes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

The distribution is unknown, so the sampling distribution will only be approximately normal when n is at least 30.

So the correct answer should be:

Approximately normal for large sample sizes

7 0
3 years ago
K is the midpoint of JL,JL=4x-2 and jk=7,find x KL and JL
diamong [38]
K is the midpoint of JL...so JK=KL=7
JL=(JK + KL)>>>>>>whole part axiom
or,4x=2×JK>>>>>>>{JK=KL➡JK+KL=JK+JK}
or,4x=2×7
or,x=14/4=7/2

[ x=7/2 ] [KL=JK=7] and [JL=4x=(4×7)/2=14]
4 0
3 years ago
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