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Salsk061 [2.6K]
3 years ago
7

the wingspan of one type of insect is 0.2 millimeter. which expression shows the numerical value of the wingspan in meters?

Mathematics
1 answer:
ozzi3 years ago
5 0

2.0 * 10^-1

hope this helps

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Is the point (83, 70) a solution to 52x - 35y = 1,866?
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Step-by-step explanation:

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You pick a card at random. You put the first card back, and you pick a second card at
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Step-by-step explanation:

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Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i
kolezko [41]

Answer:

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

Step-by-step explanation:

Data given and notation

Data:    801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=811 represent the sample mean  

s=19.647 represent the standard deviation for the sample

n=5 sample size  

\mu_o =750 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:  

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

What do you conclude?  

Compute the p-value  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

4 0
3 years ago
Write the decimal 2 as a percent
gayaneshka [121]
If 2 stands for 2 out of 10 (2/10); Then the answer is 20%.
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Read 2 more answers
Evaluate 10m + n2/4 when m = 5 and n = 4
Allushta [10]

Answer:

54

Step-by-step explanation:

we have

10m+\frac{n^{2}}{4}

For m=5.n=4

substitute the values of m and the value of n in the equation

so

10(5)+\frac{4^{2}}{4}

50+4

54

3 0
3 years ago
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