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2 years ago

Point slope form. (-5,-1) and (6,-5)

Mathematics

1 answer:

TEA [102]2 years ago

6 0

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point on a line

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have two points (-5, -1) and (6, -5). Substitute:

$m=\dfrac{-5-(-1)}{6-(-5)}=\dfrac{-5+1}{6+5}=-\dfrac{4}{11}$

Put the value of a slope and coordinates of a point to the equation of a line:

$(-5,\ -1),\ m=-\dfrac{4}{11}\\\\y-(-1)=-\dfrac{4}{11}(x-(-5))\\\\y+1=-\dfrac{4}{11}(x+5)$

$(6,\ -5),\ m=-\dfrac{4}{11}\\\\y-(-5)=-\dfrac{4}{11}(x-6)\\\\y+5=-\dfrac{4}{11}(x-6)$

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180 lb 12 oz - 79.3 kg = ___________ kg

Brrunno [24]

Answer:

$180lb\ 12 oz - 79.3kg = 2.8kg$

Step-by-step explanation:

Given

$180\ lb 12\ oz - 79.3kg$

Required

Solve

First, convert 180lb to kg

$1lb \to 0.454kg$

$180lb \to 180 *0.454kg$

$180lb \to 81.7kg$ -- to 1dp

Next, convert 12oz to kg

$1oz \to 0.03kg$

$12oz \to 12 * 0.03kg$

$12oz \to 0.4kg$ --- to 1 dp

So, the expression becomes:

$180\ lb 12\ oz - 79.3kg$

$81.7kg + 0.4kg - 79.3kg$

$= 2.8kg$

Hence:

$180lb\ 12 oz - 79.3kg = 2.8kg$

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3 years ago

Denny had five kittens (K) that he wants to sell for $15.00 each.Tge cost to raise the kittens so far has been $10.00 each. What

madreJ [45]

Five kittens will give Denny $75.
Raising them cost $50.
So Denny will make a profit of $15.

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3 years ago

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A cylinder has a circular base with a radius that measures 9 inches. Determine the diameter of the base of the cylinder. 3 inche

yarga [219]

It is 18 because the radius is half the diameter and 9 x 2 = 18.

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3 years ago

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Of the water lilies in the pond, 43% are yellow. The others are white. A frog randomly jumps onto a lily. What is the complement

stealth61 [152]

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2 years ago

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating

Natali [406]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565$

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

6 0

2 years ago