All categories

3 years ago

What is the solution w + 1 < -3

Mathematics

1 answer:

likoan [24]3 years ago

6 0

$w+1\ \textless \ -3\\
w\ \textless \ -4$

You might be interested in

Please help can complete this analogy

Goshia [24]

Hanukkah, because that is what holiday it is used for. (Though I'm Christian and I don't know what the heck a creche is) but it says Christmas, so yeah. Hanukkah.

8 0

3 years ago

How do you simplify -3x²ym + 7x - 5ymx² + 16x

brilliants [131]

you cant add or subtract or multiply or divide if the variables arent the same so the simplified version of -3x²ym + 7x - 5ymx² + 16x would be
 -3x²ym - 5ymx² + 23x

3 0

3 years ago

Read 2 more answers

dvd cases are sold in packages of 9 padded mailing envelopes are sold in packets of 3 what is the least number of cases and enve

Lostsunrise [7]

Im lost at this sorry, hope sum1 helps you out on this tho!

7 0

3 years ago

Chan hee bought 3.4 pounds that cost 6.95 per pound how much did he spend on coffee? sl

Aleks [24]

1 pound of coffee costs 6.95
3.4 pounds will cost 6.95 x 3.4 = 23.63

8 0

3 years ago

Read 2 more answers

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