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2 years ago

When 0.5(20x – 50y + 36) – 0.25(–100x + 40y – 12) is simplified, what is the resulting expression?

Mathematics

1 answer:

stiks02 [169]2 years ago

6 0

Answer:

solution,

$0.5(20x - 50y + 36) - 0.25( - 100x + 40y - 12) \\ = 10x - 25y + 18 + 25x - 10y + 3 \\$

Combine the like terms:

$= 10x + 25x - 25y - 10y + 18 + 3 \\$

Simplify:

$35x - 35y + 21$

Hope this helps...

Good luck on your assignment...

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As Saturn revolves around the sun, it travels at a speed of approximately 6 miles per second. Convert this speed to miles per mi

leva [86]

Answer:

The answer is 21,600. If it travels 6 miles per second that in a minute it would travel 360 miles. You multilpy 360 times 6 and you get your answer of 21,600 miles.

Step-by-step explanation:

5 0

2 years ago

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab

Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected, and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, $H_0$ : $\mu$ = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, $H_1$ : $\mu$ < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

s = sample standard deviation = 38.2 hours

n = sample size = 50

So, test statistics = $\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }$ ~ $t_4_9$

= -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0

2 years ago

How do you do this? Please help

PtichkaEL [24]

Answer:

$PO=86.2\\ OM=23.8\\$

Step-by-step explanation:

To preface, your figure is going to be a line segment, with $O$ as your midpoint, in between points $P$ & $M.$

With that being said:

$PO+OM=PM$

Identify your values:

$PO=7y+12\\OM=3y-8\\PM=110$

Substitute the values into the first equation:

$7y+12+3y-8=110$

Combine like terms:

$10y+4=110$

Subtract $4$ from both sides of the equation:

$10y=106$

Divide by the coefficient of $y$ , which is $10$ :

$y=10.6$

Substitute $10.6$ for $y$ in segments $PO$ & $OM$ :

$PO=7(10.6)+12$

$OM=3(10.6)-8$

$PO=74.2+12$

$OM=31.8-8$

Solve:

$PO=86.2$

$OM=23.8$

Check your answers by substituting:

$PO+OM=PM$

$86.2+23.8=110$

8 0

2 years ago

The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginnin

solniwko [45]

Answer:

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the the vines infested using Pernod 5 and vines infested using Action. The population proportion of the vines infested using Pernod 5 and vines infested using Action would be p1 and p2 respectively.

p1 - p2 = difference in the proportion of the vines infested using Pernod 5 and vines infested using Action.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

Ha : p1 ≠ p2

p1 - p2 ≠ 0

it is a two tailed test

Sample proportion = x/n

Where

x represents number of success(number of complaints)

n represents number of samples

For vines infested using Pernod 5,

x1 = 26

n1 = 410

p1 = 26/410 = 0.063

For vines infested using Action,

x2 = 39

n2 = 400

P2 = 39/400 = 0.098

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (26 + 39)/(410 + 400) = 0.08

1 - pc = 1 - 0.08 = 0.92

z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)

z = (0.063 - 0.098)/√(0.08)(0.92)(1/410 + 1/400) = - 0.035/0.019066

z = - 1.84

Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.033

We would double this area to include the area in the right tail of z = 1.84 Thus

p = 0.033 × 2 = 0.066

By using the p value,

Since alpha, 0.01 < than the p value, 0.066, then we would fail to reject the null hypothesis.

Therefore, At a 1% level of significance, we cannot conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action

5 0

2 years ago

Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us

grigory [225]

Answer:

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

$\bar X = \frac{62.532+76.478}{2}= 69.505$

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

$\bar X = \frac{62.532+76.478}{2}= 69.505$

So then the mean for this case is 69.505

5 0

2 years ago