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2 years ago

Graph y+3=5(x-4) please quickly

Mathematics

1 answer:

Lisa [10]2 years ago

3 0

Answer:

It crosses through 0,-23) and (4.6,0)

Step-by-step explanation:

You might be interested in

A national survey of companies included a question that asked whether the company had at least one bilingual telephone operator.

nirvana33 [79]

Answer:

The first option is correct. Option A is correct.

LCL = 0.270, and UCL = 0.397

80% Confidence interval = (0.270, 0.397)

Step-by-step explanation:

The data for Y and N for the 90 companies is attached to this solution provided.

Y represents companies with at least 1 bilingual operator and N represents companies with no bilingual operator.

The number of Y in the data = 30

Hence, sample proportion of companies with at least one bilingual operator = (30/90) = 0.3333

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = 0.3333

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 80% confidence level for sample size of 90 is obtained from the z-tables.

Critical value = 1.280

Standard error of the mean = σₓ = √[p(1-p)/n]

p = sample proportion

n = sample size = 90

σₓ = √[0.3333×0.6667)/90] = 0.0496891568 = 0.04969

80% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.3333 ± (1.28 × 0.04969)

CI = 0.3333 ± 0.0636021207

80% CI = (0.2696978793, 0.3969021207)

80% Confidence interval = (0.270, 0.397)

Hope this Helps!!!

3 0

3 years ago

What are the solutions to the quadratic equation -2x^2 + 6x + 3 = 0?

mario62 [17]

For this, we will be using the quadratic formula, which is $x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$ , with a=x^2 coefficient, b=x coefficient, and c = constant. Our equation will look like this: $x=\frac{-6+/-\sqrt{6^2-4*(-2)*3}}{2*(-2)}$

Firstly, solve the multiplications and the exponents: $x=\frac{-6+/-\sqrt{36+24}}{-4}$

Next, do the addition: $x=\frac{-6+/-\sqrt{60}}{-4}$

Next, your equation will be split into two: $x=\frac{-6+\sqrt{60}}{-4},\frac{-6-\sqrt{60}}{-4}$ . Solve them separately, and your answer will be $x=-0.436,3.436$

5 0

3 years ago

A pharmacist wants to make 36ml of a 65% hydrogen peroxide solution. how many pls of 75% hydrogen peroxide must he mix with 57%

faltersainse [42]

Answer:

16 mls of the 75% solution and 20 mls of the 0.57% solution.

Step-by-step explanation:

Set up a system of equations:

Let x = volume of 75% solution and y be volume of 57% solution

0.75x + 0.57y = 36*0.65

0.75x + 0.57y = 23.4.......(1)

x + y = 36.........(2)

From equation (2):

y = 36 - x

Substituting this into equation(1):

0.75x + (0.57(36 - x) = 23.4

0.75x + 20.52 - 0.57x = 23.4

0.18x = 23.40 - 20.52 = 2.88

x = 2.88/0.18

x = 16

so from equation (2): y = 36-16 = 20.

8 0

2 years ago

What is wrong in this

il63 [147K]

Step 3 is wrong ( = 1/10 pound x 1/10 pound )

7 0

3 years ago

Find the average rate of change of the function over the given interval

sattari [20]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }$

8 0

3 years ago