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3 years ago

HELP ASAP!!! EASY POINTS!!!!

Mathematics

1 answer:

RoseWind [281]3 years ago

8 0

About Point Slope Form:

Y -Y1 = m (x - X1)

m is the slope

Y1 & X1 is a point on the line

The form allows you to identify the slope & the point on the line

Y - 2 = 3 (x - -1)

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1. If y + 1/y = 2 cos teta find y

k0ka [10]

Answer:

Step-by-step explanation:

1+1/1=2

8 0

3 years ago

Read 2 more answers

Cat face x+4/5=45 sa il aflu pe x

love history [14]

Given:

The equation is

$x+\dfrac{4}{5}=45$

To find:

The value of x.

Solution:

We have,

$x+\dfrac{4}{5}=45$

Isolating the variable, we get

$x=45-\dfrac{4}{5}$

$x=\dfrac{225-4}{5}$

$x=\dfrac{221}{5}$

$x=44.2$

Therefore, the value of x is 44.2.

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3 years ago

Evaluate 27x1/3^3 I need it in simple form

ruslelena [56]

Step-by-step explanation:

$27 \times \frac{1}{ {3}^{3} }$

$= 27 \times \frac{1}{27}$

$= 1$

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2 years ago

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Help! Thank you so much!!

AveGali [126]

Your final answer is -8

5 0

3 years ago

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages ar

Elenna [48]

Answer:

We conclude that the mean amount packaged is equal to 8.17 ounces.

Step-by-step explanation:

We are given that in a particular sample of 50 packages, the mean amount dispensed is 8.171 ounces, with a sample standard deviation of 0.052 ounces.

Let $\mu$ = population mean amount packaged.

So, Null Hypothesis, $H_0$ : $\mu$ = 8.17 ounces {means that the mean amount packaged is equal to 8.17 ounces}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 8.17 ounces {means that the mean amount packaged is different from 8.17 ounces}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount dispensed = 8.171 ounces

s = sample standard deviation = 0.052 ounces

n = sample of packages = 50

So, the test statistics = $\frac{8.171-8.17}{\frac{0.052}{\sqrt{50} } }$ ~ $t_4_9$

= 0.1359

The value of t-test statistics is 0.1359.

Also, the P-value of test-statistics is given by;

the meaning of the p-value is that the p-value is the probability of obtaining a sample mean that is equal to or more extreme than 0.001 ounces away from8.17 if the null hypothesis is true.

P-value = P( $t_4_9$ > 0.136) = More than 40% {from the t-table}

Since the P-value of our test statistics is more than the level of significance of 0.01, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean amount packaged is equal to 8.17 ounces.

6 0

3 years ago