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tekilochka [14]
2 years ago
8

Evaluate for x = 3 and y = 4: x^-2y^2/x^0y^3 1/36 4/9 1024/9

Mathematics
1 answer:
GrogVix [38]2 years ago
4 0

Answer:

\frac{1}{36}

Step-by-step explanation:

Evaluating the numerator and denominator before evaluating

Using the rule of exponents

a^{-n} ⇔ \frac{1}{a^{n} }, a^{0} = 1

Thus

x^{-2}y² ← substitute values into expression

= 3^{-2} × 4² = \frac{1}{9} × 16 = \frac{16}{9}

and denominator

x^{0}y^{3} = 1 × 4³ = 1 × 64 = 64

Now dividing

\frac{16}{9} ÷ 64

= \frac{16}{9} × \frac{1}{64} ( cancel 16 and 64 )

= \frac{1}{9(4)} = \frac{1}{36}

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Svet_ta [14]

1.981: One and nine hundred eighty-one thousandths

6 0
2 years ago
The American Water Works Association reports that the per capita water use in a single-family home is 63 gallons per day. Legacy
Debora [2.8K]

Answer:

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

df=n-1=28-1=27  

p_v =P(t_{(27)}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=60 represent the sample mean

s=8.9 represent the sample standard deviation

n=28 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 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.  

We need to conduct a hypothesis in order to check if the true mean is less than 63, the system of hypothesis would be:  

Null hypothesis:\mu \geq 63  

Alternative hypothesis:\mu < 63  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=28-1=27  

Since is a one side lower test the p value would be:  

p_v =P(t_{(27)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

5 0
3 years ago
Based on the given information, which conjecture is invalid?
a_sh-v [17]
Valid based on the given information<span>. If not, write </span>invalid<span>. ... So, the statement is </span>invalid<span>. Determine .....disprove a </span>conjecture<span> reached using inductive or deductive</span><span> .</span>
3 0
3 years ago
A random sample of 10 single mothers was drawn from a Obstetrics Clinic. Their ages are as follows: 22 17 27 20 23 19 24 18 19 2
mel-nik [20]

Answer:

0.09

Step-by-step explanation:

Let x = ages of mother

x  :  22   17    27    20    23     19      24    18    19    24

N = 10

Mean = ∑x/N = 218/10 = 21.8

Difference in mean = 21.8 - 20 = 1.8

If significance level = 5% or 0.05

∴ Rejection region = 1.8 X 0.05 = 0.09

8 0
3 years ago
What is the circumference of a circle with a diameter of 45 centimeters
Free_Kalibri [48]
Keeping in mind the formula for circumference is Pi x Diameter, you simply multiply 45 by pi.
This gives you 141.3716694
4 0
3 years ago
Read 2 more answers
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