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Bogdan [553]
3 years ago
9

Rround 0.023 to nearest hund

Mathematics
2 answers:
Eva8 [605]3 years ago
6 0
0.02 thts the answer because and thing below four u round down any thing above four you round up
makkiz [27]3 years ago
5 0
When rounding to the nearest hundredth, you must look to the thousandths place. since it is a 3,  it rounds down, to **0.02**
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Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065. New equipment is being
Ilya [14]

Answer:

Yes, the new equipment appear to be effective in reducing the variation of​ weights.

Step-by-step explanation:

We are given that Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065.

A simple random sample of 25 quarters is obtained from those manufactured with the new​ equipment, and this sample has a standard deviation of 0.047.

Let \sigma = <u><em>standard deviation of weights of new equipment.</em></u>

SO, Null Hypothesis, H_0 : \sigma \geq 0.065      {means that the new equipment have weights with a standard deviation more than or equal to 0.065}

Alternate Hypothesis, H_A : \sigma < 0.065      {means that the new equipment have weights with a standard deviation less than 0.065}

The test statistics that would be used here <u>One-sample chi-square</u> test statistics;

                           T.S. =  \frac{(n-1)s^{2} }{\sigma^{2} }  ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.047

           n = sample of quarters = 25

So, <u><em>the test statistics</em></u>  =  \frac{(25-1)\times 0.047^{2} }{0.065^{2} }  ~  \chi^{2}__2_4   

                                     =  12.55

The value of chi-square test statistics is 12.55.

Now, at 0.05 significance level the chi-square table gives critical value of 13.85 at 24 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of chi-square as 12.55 < 13.85, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that the new equipment have weights with a standard deviation less than 0.065.

5 0
3 years ago
Find f(6) if f(x) = x2 ÷ 3 + x. <br><br> 1.) 4 <br> 2.) 10 <br> 3.) 18
rodikova [14]
Plug 6 for x.

f(6) = 6^2 / 3 + 6
f(6) = 36 / 3 + 6
f(6) = 12 + 6
f(6) = 18
5 0
3 years ago
If gas costs $2.19 per gallon, how many whole gallons of gas could you buy with $17.00? (Make sure to answer with a whole number
ankoles [38]

Answer:

2.19*17.00

=37.23

=$37

Step-by-step explanation:

4 0
2 years ago
What is a=16 b= 28 and c=? What would c be?
sergiy2304 [10]
A^2+b^2= c^2
16^2+ 28^2= c^2
 square root of 256+784 = square root of 1,040
32.25=32.25

C= 32.25
3 0
3 years ago
Write a x (-a)x 13 x a x (-a) x 13 in power notation
miss Akunina [59]

Answer:

a\times (-a)\times 13\times a\times (-a)\times 13 can be written in power notation as a^{4}\times 13^{2}

Step-by-step explanation:

The given expression

a\times (-a)\times 13\times a\times (-a)\times 13

Writing a\times (-a)\times 13\times a\times (-a)\times 13 in power notation:

Let

a\times (-a)\times 13\times a\times (-a)\times 13

= [13\times13][(a\times (-a)\times a\times (-a)]

As

13\times13 = 13^{2} , a\times a = a^{2} , (-a)\times (-a) = (-a)^{2}

So,

=[13^{2}][a^2\times (-a)^2]

As

(-a)^2 = a^{2}

So,

=[13^{2}][a^2\times a^2]

As ∵a^{m} \times a^{n}=a^{m+n}

=[13^{2}][a^{2+2}]

As ∵a^{m} \times a^{n}=a^{m+n}

=13^{2}\times a^{4}

=a^{4}\times 13^{2}

Therefore, a\times (-a)\times 13\times a\times (-a)\times 13 can be written in power notation as a^{4}\times 13^{2}

<em>Keywords: power notation</em>

<em>Learn more about power notation from brainly.com/question/2147364</em>

<em>#learnwithBrainly</em>

5 0
3 years ago
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