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Anvisha [2.4K]
3 years ago
6

Find 32% of 35.? plzz help me with this

Mathematics
2 answers:
Anna35 [415]3 years ago
7 0
32% of 35 is 32% times 35, which is 11.2
Rainbow [258]3 years ago
4 0
<span>32% of 35

First, change the percentage into a decimal. This is done by moving the decimal point two points to the left.
32% = 0.32

0.32 of 35

Whenever you see the word "of", immediately think of multiplication. Replace "of" with "</span>×".
0.32 × 35 = 11.2

So the answer is 11.2.
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Answer:

3/500

Step-by-step explanation:

42/7000 simplified is 3/500 or in decimal form it's 0.006

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A seven-sided number cube with sides numbered 1-7 is rolled. what is the probability that a 4 or an odd number is rolled.
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The sides of this rectangle are
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The longer side is at first 7cm and the shorter is 2cm.

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What’s bigger 10 to the -11 power or 10 to the -9 power ?
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Step-by-step explanation:

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3 years ago
A study of pollutants showed that certain industrial emissions should not exceed 2.3 parts per million. You believe a particular
Musya8 [376]

Answer:

t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5    

The degrees of freedom are given by:

df=n-1=9-1=8  

And the p value would be:

p_v =P(t_{(8)}>7.5)=0.0000346  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

Step-by-step explanation:

Information provided

\bar X=3.3 represent the sample mean in ppm

s=0.4 represent the sample standard deviation

n=9 sample size  

\mu_o =2.3 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to verify if the true mean is higher than 2,3 ppm, the system of hypothesis would be:  

Null hypothesis:\mu \leq 2.3  

Alternative hypothesis:\mu > 2.3  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5    

The degrees of freedom are given by:

df=n-1=9-1=8  

And the p value would be:

p_v =P(t_{(8)}>7.5)=0.0000346  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

6 0
3 years ago
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