All categories

3 years ago

Solve the linear equation 2.25 – 11j – 7.75 + 1.5j = 0.5j – 1.

Mathematics

2 answers:

Citrus2011 [14]3 years ago

8 0

Answer:

Option (a) is correct.

2.25 – 11j – 7.75 + 1.5j = 0.5j – 1 simplifies to j = -0.45

Step-by-step explanation:

Given : The linear equation 2.25 – 11j – 7.75 + 1.5j = 0.5j – 1.

We have to solve the given linear equation.

Consider the given linear equation 2.25 – 11j – 7.75 + 1.5j = 0.5j – 1.

Subtract 0.5j both side, we have,

2.25 – 11j – 7.75 + 1.5j - 0.5j = 0.5j – 1 - 0.5j

Simplify, we have,

2.25 – 11j – 7.75 + 1.5j - 0.5j = -1

Adding like terms, we have,

-5.5 -10j = -1

Adding 5.5 both side, we have,

-10j = -1 + 5.5

-10j = 4.5

Divide both side by -10, we have,

j = -0.45

Thus, 2.25 – 11j – 7.75 + 1.5j = 0.5j – 1 simplifies to j = -0.45

disa [49]3 years ago

3 0

The correct answer is:
j= -0.45

You might be interested in

51 dollars is 75 percent of the original price. what is the original price?

makvit [3.9K]

89.25 is the original price first i did 75÷100=0.75×51=38.25+51=89.25 is the correct answer.

5 0

3 years ago

What is the value of 9 in 89,170,326

Pie

Nine Million 9,000,000

3 0

3 years ago

Read 2 more answers

D=1/2at² for solve for a

rjkz [21]

Answer:

a = d/0.5t²

divide both sides by 1/2t²

4 0

3 years ago

Read 2 more answers

Find an equation of the tangent line to the bullet-nose curve y=|x|/sqrt(2−x^2) at the point (1,1) I think that square root is w

Mashcka [7]

$\bf y=\cfrac{|x|}{\sqrt{2-x^2}}\qquad \boxed{|x|=\pm\sqrt{x^2}}\qquad y=\cfrac{\sqrt{x^2}}{\sqrt{2-x^2}}\\\\
-------------------------------\\\\
\cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x\cdot \sqrt{2-x^2}~~-~~\sqrt{x^2}\cdot \frac{1}{2}(2-x^2)^{-\frac{1}{2}}\cdot -2x}{(\sqrt{2-x^2})^2}}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{x\sqrt{2-x^2}}{\sqrt{x^2}}~~+~~\frac{x\sqrt{x^2}}{\sqrt{2-x^2}}}{2-x^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{x(2-x^2)~~+~~x(x^2)}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{2x\underline{-x^3+x^3}}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}$

$\bf \cfrac{dy}{dx}=\cfrac{2x}{\sqrt{x^2}\sqrt{2-x^2}(2-x^2)} \implies \boxed{\cfrac{dy}{dx}=\cfrac{2x}{|x|\sqrt{2-x^2}(2-x^2)}}
\\\\\\
\left. \cfrac{dy}{dx} \right|_{1,1}\implies \cfrac{2(1)}{|1|\cdot \sqrt{2-1^2}(2-1^2)}\implies 2\\\\
-------------------------------\\\\
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=2(x-1)\implies y-1=2x-2
\\\\\\
y=2x-1$

5 0

3 years ago

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed

tatyana61 [14]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 8 minutes

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes

(b) The P-value of the test statistics is 0.0436.

(c) We conclude that the actual mean waiting time equals the standard at 0.05 significance level.

(d) 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Step-by-step explanation:

We are given that the length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes.

A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minutes.

Let $\mu$ = actual mean waiting time.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 8 minutes {means that the actual mean waiting time does not differs from the standard}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes {means that the actual mean waiting time differs from the standard}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

So, test statistics = $\frac{8.5-8}{\frac{3.2}{\sqrt{120} } }$

= 1.71

The value of t test statistics is 1.71.

(b) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.9564 = 0.0436

(c) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual mean waiting time equals the standard.

(d) Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population proportion, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $8.5-1.96 \times{\frac{3.2}{\sqrt{120} } }$ , $8.5+1.96 \times{\frac{3.2}{\sqrt{120} } }$ ]

= [7.93 minutes , 9.07 minutes]

Therefore, 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Yes, it support our above conclusion.

3 0

3 years ago