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3 years ago

A snake slithers 2/9 mile in 4/5 hour.

Mathematics

2 answers:

kykrilka [37]3 years ago

8 0

Answer:

Your answer is A. 8/45 mph

Step-by-step explanation:

I can provide a more detailed answer in a few but for starters, you can rule out answers C & D because that would be the fastest snake alive to be able to travel that far in such a short time.

That leaves a choice between A or B. Using the common denominator method to switch 2/9 and 4/5 to x/45, this makes the most sense. When comparing the 8/45 mph to the actual question, it's very reasonable.

I'll come back in a few and provide the actual math. I'm late for the doctor.

Good luck! Hope this helps!

9966 [12]3 years ago

6 0

Answer:

B: 5/18 miles per hour

Step-by-step explanation:

1. multiply the fractions by the reciprocal

2/9 × 4/5 = 10/36

2. Divide answer by common multiple which is 2

10/36 ÷ 2 = 5/18

5/18 miles per hour is the snakes speed

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3 years ago

If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g

lara [203]

Answer:

$p_v =P(t_{(n-1)}>t_{calculated}) =0.0601$

The p value on this case is given by the problem.

If we compare the p value with a significance level assumed $\alpha=0.05$ , we see that $p_v > \alpha$ and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value before , y = test value after

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

The first step is calculate the difference $d_i=y_i-x_i$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=t_{calculated}$

The next step is calculate the degrees of freedom given by:

$df=n-1$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

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4 0

3 years ago

A country's population in 1991 was 114 million. In 1997 it was 120 million. Estimate the population in 2014 using exponential gr

Rufina [12.5K]

The answer is 138 million.

Step 1. Calculate the growth rate.
Step 2. Calculate the population number in 2014.

We will use the formula for exponential growth:
A = P * eⁿˣ
where:
A - the final amount (value)
P - the initial amount (value)
e - the mathematical constant (e ≈ 2.72)
n - the growth rate
x - the time

Step 1:
A = 120 million = 120 000 000
P = 114 million = 114 000 000
n = ?
x = 1997 - 1991 = 6
Therefore:
$120 000 000 = 114 000 000 * e ^{n*6} \\ e ^{n*6} =120 000 000 /114000000 \\ e ^{n*6} =1.05 \\$
Logarithm both sides of the equation:
$ln(e ^{n*6})=ln(1.05) \\ n*6*ln(e) = ln(1.05) \\ 6n=ln(1.05) \\ n = \frac{ln(1.05)}{6} \\ n = \frac{0.05}{6} \\ n = 0.0083$

Step 2:
Now when we know the growth rate, it is easy to estimate the population in 2014
A = ?
P = 120 000 000
n = 0.0083
t = 2014 - 1997 = 17
$A = 120 000 000 * e ^{0.0083*17} \\ 
A = 120 000 000 * e^{0.1411} \\ 
A = 120 000 000 * 1.15 \\ 
A = 138 000 000
$
Thus, the population will be 138 million in 2014

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3 years ago