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3 years ago

Is it illegal in california to eat raw oranges in bathroom?

Mathematics

2 answers:

viktelen [127]3 years ago

4 0

Answer:

nah it's illegal in cali to eat raw oranges in the bathtub, but im not sure if that's a real law

so i guess if ur on the toilet or smth you can eat it...but like why would you? like in the bathroom?

iren2701 [21]3 years ago

4 0

Answer:

No, it's not.

Step-by-step explanation:

It's not even a real law. People used to think that the citric acid coming from the orange would mix with bath oils and create an explosion. As it turns out, citric acid, obviously, does not cause any explosions when mixed with bath oils. It's just gossip.

You might be interested in

Use number line to compare the numbers

denis23 [38]

Answer:

0 > -60

Step-by-step explanation:

Every negative number is less than 0

6 0

3 years ago

Cam returned 8 bottles worth 5 cent each and 5 bottles worth 10 cent each. How much money did he get?

vlabodo [156]

Answer:

He received 90 cents for the bottles.

Step-by-step explanation:

8 bottles × 5¢ = 40¢

5 bottles × 10¢ = 50¢

40¢ + 50¢ = 90¢

4 0

4 years ago

A poll surveyed 503 video gamers, and 106 of them said they prefer playing games on a console rather than a computer. An executi

Kitty [74]

Answer:

The claim has no evidence to be supported.

On the contrary, there is enough evidence to support the claim that <em>less than 25% of gamers prefer consoles</em>.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The sample has a size n=503.

The sample proportion is p=0.211.

p=X/n=106/503=0.211

The sample proportion is less than 25%, so we will test the claim that <em>less </em>than 25% of gamers prefer consoles.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.25\\\\H_a:\pi$

The significance level is assumed to be 0.05.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.25*0.75}{503}}\\\\\\ \sigma_p=\sqrt{0.000373}=0.019$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.211-0.25+0.5/503}{0.019}=\dfrac{-0.038}{0.019}=-1.9685$

This test is a left-tailed test, so the P-value for this test is calculated as:

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

As the P-value (0.0245) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that less than 25% of gamers prefer consoles.

3 0

3 years ago

According to a study by the American Pet Food Dealers Association, 63 percent of U.S. households own pets. A report is being pre

Elis [28]

Answer: 2.511 lies in the rejection region, so we will reject the null hypothesis.

jhgjhghjh

Step-by-step explanation:

Since the null and alternate hypothesis are :

$H_0:\pi =0.63\\\\H_1:\pi \neq 0.63$

Since there is 0.05 level of significance,

So, there will be two tail test:

As we know that

z = -1.96 or z = 1.96

If the test statistic does not fall in the range of -1.96 to 1.96, we will reject the null hypothesis.

So, it becomes,

$z=\dfrac{p-\pi }{\sqrt{\dfrac{\pi(1-\pi)}{n}}}\\\\z=\dfrac{0.7-0.63}{\sqrt{\dfrac{0.63.\times 0.37}{300}}}\\\\z=2.511$

Since 2.511 lies in the rejection region, so we will reject the null hypothesis.

Hence, People does not seem to be representative of the Pet Food survey respondents.

6 0

3 years ago

On the first day of June, there were about 17.71 h of daylight in a city. Five months later, there were about 5.30 h of daylig

Andru [333]

Answer:

Percentage increases and decreases, are found by setting the problem up proportionally.

You are trying to find what percentage of 17.37 is 8.30 first.

so What %/100 = 8.30/17.37

Now define a variable:

Let x = What%

x/100 = 8.30/17.37

17.37x = 830

x = 830/17.37

x ≈ 47.78

8.30 is approximately 47.78% of 17.37, so the percentage decrease is 100% - 47.78% = 52.22%

The percentage decrease is approximately 52.22%

7 0

3 years ago