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3 years ago

PLEASE NEED HELP ASAP

Mathematics

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

reflect over the y axis, move two units to the left and 8 units down

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Select all that apply. Which of the following expressions have a quotient of -1/7?

Strike441 [17]

Answer:

we need more

Step-by-step explanation:

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3 years ago

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Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$

The t-value for a 95% confidence interval is t=1.96.

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

$MOE=t\cdot s_M=1.96 \cdot 475.795=932.54$

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

$LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13$

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

6 0

3 years ago

Select the correct answer.

Alex787 [66]

Answer:

A) y=3/2x-12

Step-by-step explanation:

Slope intercept formula is y=mx+b

mx is the slope.

B is the y intercept

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Which word is an antonym of permit?

Deffense [45]

Answer:

What are the options? ban, forbid, prohibit, are some examples

Step-by-step explanation:

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3 years ago

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago