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3 years ago

What does it mean when a guy that you text gets mad at you because your shirt shows your stomach?

2 answers:

8 0

It could mean that he’s being overprotective of you and cares about what u wear with certain people or during certain occasions.

mariarad [96]3 years ago

7 0

Answer:

depends on how you look at it. It means that they either think your fat and your shirt is too small or that your being too revealing and need to cover up your stomach.

Explanation:

Once before a guy got mad at me because of the same reason but it was only because he cared for my privacy and protection :)

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Nylon is an important component in many products. One step in the production of nylon is the reaction of cyclohexane with oxygen

UkoKoshka [18]

Answer:

Follows are the answer to this question:

Explanation:

pleas find the attachment file.

Formula for ΔE:

$\Delta E = \sum \text{Bond energy of recatants}- \sum \text{Bond energy of Products}$

$= 6 ( B.E_{C-C}) +12 (B.E_{C-H}) +(B.E_{o \ = \ o}) - 6(B.E_{C-C})-10(B.E_{C-H}) -( B.E_{C=0}) -2(B.E_{0-H})\\\\ = \bold{2 (B.E_{C-H}) +(B.E_{o \ = \ o}) -( B.E_{C=0}) -2(B.E_{0-H})}\\\\$

to calculate the value for ΔE put the value in above formula:

4 0

3 years ago

What Happens When extra Solute is added to unsaturated ,saturated and supersaturated solution

Ymorist [56]

of water is 36.0 g. If any more NaCl is added past that point, it will not dissolve because the solution is saturated. If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated.

8 0

3 years ago

HELP HELP HELP HELP!!!!!!

suter [353]

Answer:

I think it is 100 if I guess

8 0

3 years ago

9. The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aque

Nezavi [6.7K]

This question is incomplete, the complete question is;

The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aqueous film-forming foam is as follows; 27 41 22 27 23 35 30 33 24 27 28 22 24

( see " use of AFFF in sprinkler systems," Fire technology, 1975: 5)

The system has been designed so that the true average activation time is supposed to be at most 25 seconds.

Does the data indicate the design specifications have not been met?

Test the relevant hypothesis at significance level 0.05 using the P-value approach

Answer:

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

Explanation:

Given the data in the question;

lets consider Null and Alternative hypothesis;

Null hypothesis H₀ : There is sufficient evidence that the true average activation time is at most 25 seconds

Alternative hypothesis H₁ : There is no sufficient evidence that the true average activation time is at most 25 seconds

i.e

Null hypothesis H₀ : μ ≤ 25

Alternative hypothesis H₁ : μ > 25

level of significance σ = 0.05

first we determine the sample mean;

$x^{bar}$ = $\frac{1}{n}$ ∑ $x_{i}$

where n is sample size and ∑ $x_{i}$ is summation of all the sample;

= $\frac{1}{13}$ ( 27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24 )

= $\frac{1}{13}$ ( 363

sample mean $x^{bar}$ = 27.9231

next we find the standard deviation

s = √( $\frac{1}{n-1}$ ∑( $x_{i}-x^{bar}$ )²

x ( $x_{i}-x^{bar}$ ) ( $x_{i}-x^{bar}$ )²

27 -0.9231 0.8521

41 13.0769 171.0053

22 -5.9231 35.0831

27 -0.9231 0.8521

23 -4.9231 24.2369

35 7.0769 50.0825

30 2.0769 4.3135

33 5.0769 25.7749

24 -3.9231 15.3907

27 -0.9231 0.8521

28 0.0769 0.0059

22 -5.9231 35.0831

24 -3.9231 15.3907

sum 378.9229

so ∑( $x_{i}-x^{bar}$ )² = 378.9229

∴

s = √( $\frac{1}{13-1}$ ×378.9229 )

s = √31.5769

standard deviation s = 5.6193

now, the Test statistics

t = ( $x^{bar}$ - μ ) / $\frac{s}{\sqrt{n} }$

we substitute

t = ( 27.9231 - 25 ) / $\frac{5.6193}{\sqrt{13} }$

t = 2.9231 / 1.5585

t = 1.88

now degree of freedom df = n - 1 = 13 - 1 = 12

next we calculate p-value

p-value = 0.042299 ( using Execl's ( = TDIST(1.88,12,1)))

Here x=1.88, df=12, one tail

now we compare the p-value with the level of significance

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

3 0

3 years ago

As the temperature of a gas increases, explain what happens to its density and why.

Ghella [55]

Answer:

When a gas is heated, the molecules move faster, bump into each other, and spread apart. Because the molecules are spread apart, they take up more space. They are less dense.

3 0

2 years ago

Read 2 more answers