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3 years ago

what is the y intercept of the line with a slope of - 4/7 that also passes through the point (7 , - 7/2) ?

Mathematics

1 answer:

kondor19780726 [428]3 years ago

8 0

The answer is the last option: $\frac{1}{2}$

Explanation:

1. You have that the equation of the line is:

$y=mx+b$

Where $m$ is the slope of the line and $b$ is the intercept with the y-axis.

2. Therefore, you must find $b$ .

3. Then, you must substitute the points given in the problem and the slope into the equation of the line and solve for $b$ , as following:

$y=mx+b\\-\frac{7}{2}=(-\frac{4}{7})(7)+b\\-\frac{7}{2}=-4+b\\b=\frac{1}{2}$

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a man sells an article at a gain of 15%. if it was sold for rupees 600 he would have gained 20%. find the selling price of the a

DanielleElmas [232]

First, you multiply 600 and .20 to find 20% of 600 rupees, which is 120 rupees. Second, you subtract the 20%, (120 rupees) from 600 rupees, which is 480 rupees. After that, you have the original selling price of 480 rupees that the man purchased the article for, but the question is stating that he gained 15% of the original sales price. So, you multiply 480 and .15 to find 15% of 480 rupees, which is 72. However, he GAINED more money than what he payed for the article so you add 72 rupees to 480 rupees to get 552 rupees. In conclusion, the man sells the article for 552 rupees.

4 0

2 years ago

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)

skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Step-by-step explanation:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample x⁻ = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

Step 2:-

The 95% of confidence intervals for the average spending

( $(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

$(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )$

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{28-30 }{\frac{10}{\sqrt{25} } }$

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

Conclusion:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0

2 years ago

What is 7.4% written as a decimal?

BlackZzzverrR [31]

$p\%=\frac{p}{100}\\\\7.4\%=\frac{7.4}{100}=\frac{7.4\cdot10}{100\cdot10}=\frac{74}{1000}=\boxed{0.074}$

5 0

3 years ago

Please help me ill give you brianlest

Neko [114]

Answer:

A convex mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the right of the mirror.

plz can i get brainliest:)

7 0

2 years ago

Read 2 more answers

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

2 years ago