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lys-0071 [83]
3 years ago
14

What is the percent of 1.6 hours to 0.95 hour

Mathematics
2 answers:
Stolb23 [73]3 years ago
7 0
1.6*100/0.95=168%     1.6 to 0.95
0.95*100/1.6=59%       0.95 to 1.6
Akimi4 [234]3 years ago
4 0

Answer:

168.4%  of 1.6 hours to 0.95 hour.

Step-by-step explanation:

Given : 1.6 hours to 0.95 hour.

To find : what is the percent.

Solution : We have given

1.6 hours to 0.95 hour.

Percentage  = \frac{1.6}{0.95}*100.xzs

Percentage  = 1.684 * 100

Percentage=  168.4%

Therefore, 168.4%  of 1.6 hours to 0.95 hour.

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What is the solution of the system? Use the elimination method. 2x+y=206x−5y=12 . Enter your answer in the boxes.
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2x+y=206x-5y=12
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Read 2 more answers
Can someone please help me factorize this one:<br><img src="https://tex.z-dn.net/?f=0.01y%20%7B%7D%5E%7B2%7D%20%20-%20%20%5Cfrac
Scorpion4ik [409]

First let us convert 0.01 into fraction:

\frac{y^{2}}{100}  -\frac{1}{4}

Now we convert 100 and 4 into perfect squares as follows:

(\frac{y}{10})^{2}-(\frac{1}{2})^{2}

Now we use the property of :

x^{2} -y^{2} =(x+y)(x-y)

(\frac{y}{10} +\frac{1}{2} ) (\frac{y}{10} -\frac{1}{2} )

4 0
3 years ago
The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the
lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average sodium content.</em></u>

So, Null Hypothesis, H_0 : \mu = 1100 milligrams      {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, H_A : \mu \neq 1100 milligrams      {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about population standard deviation;

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

where, \bar X = sample average sodium content = 1088.64 milligrams

            s = sample standard deviation = 234.12 milligrams

            n = sample of packages of Oriental Spice Sauce = 40

So, <u><em>test statistics</em></u>  =  \frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}  ~ t_3_9

                              =  -0.307

The value of z test statistics is -0.307.

<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>

<em>Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the sodium content is same as what the nutrition label states.

3 0
3 years ago
2 over 3 divided by 5
Yuliya22 [10]
2/3 divide by 5 is 2/15

Remember keep change flip

2/3 times 1/5 is 2/15

Hope this helps!
6 0
3 years ago
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