Round 5.275 to the nearest hundredth

1 answer:

6 0

5.28....What you do is find the hundredth place then locate the number to the right of it. If the number is greater than 4, then you round the hundredth place up a number and if it is less than or equal to 4 then leave the hundredth place alone. but in this case we round it up because 5 is greater than 4 so change 7 to eight.

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Epsilon Eridani is a star that is 10.5 light years away from Earth. One light year is about 9.46 trillion kilometers.

ycow [4]

Answer:

just multiply the two numbers

10.5*9.46=99.33

Step-by-step explanation:

7 0

3 years ago

An article in Fire Technology describes the investigation of two different foam expanding agents that can be used in the nozzles

kvv77 [185]

Answer:

a) There is no evidence to support the claim that there is no difference in mean foam expansion of these two agents.

b) P=0

c) 90% CI

$-3.2406\leq \mu_1-\mu_2 \leq -2.2614$

This CI tells us that there is a 90% confidence that the real value of the difference between the means is between this two values. We see that both are negative values, so the value 0 is left out of the interval.

That means we can be almost sure both means don't have the same value, confirming the results of the previous test.

Step-by-step explanation:

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq 0$

The significance level is 0.10 and it will be a two-tailed test.

The difference of the sample means is:

$M_d=M_1-M_2=4.340-7.091=-2.751$

As the sample size is equal for both samples, the estimated standard error of the difference between means is calculated as:

$s_M_d=\sqrt{\frac{s_1^2+s_2^2}{N}}= \sqrt{\frac{0.508^2+0.430^2}{5}}=\sqrt{\frac{0.442964}{5} }= \sqrt{0.0886}=0.2976$

Then, the statistic z is:

$z=\frac{M_d-(\mu_1-\mu_2)}{s_M_d}=\frac{-2.751-0}{0.2976}=-9.24\\\\P(|z|>9.24)=0$

The P-value (P=0) is much lower than the significance level, so the null hypothesis is rejected. The means are different.

For a 90% confidence interval for the difference of the means, we use a z=1.645.

Then the confidence interval is defined as:

$M_d-z*s_M_d\leq \mu_1-\mu_2 \leq M_d-z*s_M_d\\\\-2.751-1.645*0.2976\leq \mu_1-\mu_2 \leq -2.751+1.645*0.2976\\\\-3.2406\leq \mu_1-\mu_2 \leq -2.2614$