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4 years ago

This graph shows the solution to which inequality (3 3) (-3 -1)

Mathematics

2 answers:

Norma-Jean [14]4 years ago

6 0

Answer:

<h3> $\boxed{y > \frac{2}{3} x + 1}$ </h3>

Option B is the correct option.

Step-by-step explanation:

point ₁ ( - 3 , - 1 ) x₁ = - 3 , y₂ = - 1

point ₂ ( 3 , 3 ). x₂ = 3 , y₂= 3

Now, let's find the slope:

Slope ( m ) = $= \frac{y2 - y1}{x2 - x1}$

$= \frac{3 - ( - 1)}{3 - ( - 3)}$

$= \frac{3 + 1}{3 + 3}$

$= \frac{4}{6}$

$= \frac{2}{3}$

At ( 3 , - 1 )

y = mx + c , where m is the gradient / slope amd c is called the intercept on y-axis

$- 1 = \frac{2}{3} \times ( - 3) + c$

Solve for c

$- 1 + 2 = c$

$c = 1$

Since, The red line is dotted line. Therefore it does not include the equal to part. And the shaded region is upper part. Hence, ' > ' should be used.

The answer would be:

$y > \frac{2}{3} x + 1$

Hope I helped!

Best regards!

Digiron [165]4 years ago

5 0

Answer:

Step-by-step explanation:

since the graph shows dotted line the sign has to be < or > so C and D eliminated from your answer

y>2/3 x +1 is your answer (B)

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A tire company has developed a new type of steel-belted radial tire. Extensive testing indicates the population of mileages obta

Andru [333]

Answer:

Step-by-step explanation:

According to the given question, a tire company has developed a new type of steel-belted radial tire. Extensive testing indicates the population of mileages obtained by all tires of this new type is normally distributed with a mean of 37,000 miles and a standard deviation of 3,887 miles.

Let us define X be the random variable shows that the mileages tires normally distributed with

mean

μ = 37000

standard deviation

σ =3, 887

Therefore

X ~ (μ = 37000, σ =3,887)

The company wishes to offer a guarantee providing a discount on a new set of tires if the original tires purchased do not exceed the mileage stated in the guarantee. Therefore the guaranteed mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage is determined as:

P(X < k) = 0.02

$\Rightarrow P\left ( \frac{X-\mu }{\sigma }\leq \frac{k-\mu }{\sigma } \right )=0.02
\\\\P\left ( Z \leq \frac{k-37,000 }{3,887 } \right )=0.02$

From the standard normal curve 2% area is determined as -2.0537 and hence

$\frac{k-37,000}{3,887}=-2.0537\\\\k=37000-7982.7319\\\\k=29017.2681\\\\\Rightarrow k=29018$

If we consider z value at two decimal places then

$\frac{k-37,000}{3,887} =-2.05\\\\\Rightarrow k=29032
$

Therefore the guaranteed 29032 mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage.

The area under the standard normal curve is determined as:

6 0

4 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$