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3 years ago

What is 7-4x +2 (5x+8) if x= -2 with work plsssss

1 answer:

7 0

7−4x+2((5)(2)+8)=7+−4x+36Combine Like Terms:=7+−4x+36=(−4x)+(7+36)=−4x+43Answer:=−4x+43

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Convert 4x-2y=6 into slope intercept form

DanielleElmas [232]

Answer:

y=2x-3

Step-by-step explanation:

first, you have to get the y by itself (isolate the y). you do this by subtracting 4x from both sides.

4x-2y=6

-2y=-4x+6

next, you have to divide both side by -2 to isolate the y further and get this into slope-intercept form:

-2y=-4x+6

y=2x-3

5 0

3 years ago

Find r(t) if r'(t)=ti+e^tj+te^tk and r(0)=i+j+k

weeeeeb [17]

R(t) = integral of r'(t) = integral of ti + e^tj + te^tk = 1/2t^2i + e^tj + (te^t - e^t)k + c
r(0) = j - k + c = i + j + k
c = i + 2k
Therefore, r(t) = (1/2t^2 + 1)i + e^tj + (te^t - e^t + 2)k

6 0

4 years ago

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago

Find the zeros of the function.

Verdich [7]

Smaller x = -2/5

Larger x = 5

7 0

3 years ago

Why isn't 14.5% equivalent to the decimal 14.5

Lerok [7]

$14.5\%=\frac{14.5}{100}=\frac{145}{1000}=0.145 \\ \\
0.145 \not= 14.5$

6 0

3 years ago

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