The solution to the problem is as follows:
We have 2+.8(2) + .8(.8(2)) + .8(.8(.8(2))) + ... =
2( .8^0 + .8^1 + .8^2 + .8^3 + ... ) =
2(.8^n -1) / (.8-1) . As n-->infinity, .8^n-->0 giving us
<span>2(-1)/(-.2) = 2(5) = 10 meters.
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If each lap in a pool is 100 meters long,how many laps equal one mile
Round to the nearest tenth.(Hint:1 foot=0.3048 meter)
1 mile = 5280 ft
lets do a ratio: 1ft/.3034m = x ft/100m
the ft and meters sybols cancel, so 1/.3048 =x/100
so 100/.3038 = x = 329.164
so there are 329.164 ft for every 100 meters
to find the number of laps to get to a mile which is 5280, do another ratio
329.164ft/100 m =5280 ft/xm
the left side reduces to 3.29164 =5280/x
you can compute this and see that 5280/3.29164 = 1604.064 meters
1604.064 meters *1 lap / 100m = 16.04064 laps are required to make a mile
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: height of seaweed.
X~N(μ;σ²)
μ= 10 cm
σ= 2 cm
You have to find the value of the variable X that separates the bottom 0.30 of the distribution from the top 0.70
P(X≤x)= 0.30
P(X≥x)= 0.70
Using the standard normal distribution you have to find the value of Z that separates the bottom 0.30 from the top 0.70 and then using the formula Z= (X-μ)/σ translate the Z value to the corresponding X value.
P(Z≤z)= 0.30
In the body of the table look for the probability of 0.30 and reach the margins to form the Z value. The mean of the distribution is "0" so below 50% of the distribution you'll find negative values.
z= -0.52
Now you have to clear the value of X:
Z= (X-μ)/σ
Z*σ= X-μ
X= (Z*σ)+μ
X= (-0.52*2)+10= 8.96
The value of seaweed height that divides the bottom 30% from the top 70% is 8.96 cm
I hope this helps!
Answer:
81%
Step-by-step explanation:
Let 'L' be the dominant and 'l' e the recessive allele for ‘lazybuttness’.
Since ‘lazybuttness’ is an autosomal dominant condition, the 19% of students affected by the condition correspond to the homozygous dominant (LL) and heterozygous (Ll) genotypes. Therefore, the rest of the population has the homozygous recessive genotype (ll) and is not affected. The frequency of students not affected is:
F = 100% - 19% = 81%
Answer:
For this case we want to test if the mean number of filled overnight beds is over 523. If X represent our random variable "number of filled overnight beds", the system of hypothesis are:
Null Hypothesis: 
Alternative hypothesis: 
And for this case after conduct the test is FAIL to reject the null hypothesis. So then we can conclude that the claim that the number of filled overnight beds is over 523 is not statistically supported
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Solution to the problem
For this case we want to test if the mean number of filled overnight beds is over 523. If X represent our random variable "number of filled overnight beds", the system of hypothesis are:
Null Hypothesis:
Alternative hypothesis:
And for this case after conduct the test is FAIL to reject the null hypothesis. So then we can conclude that the claim that the number of filled overnight beds is over 523 is not statistically supported
