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grigory [225]
3 years ago
12

1. Use the clues below to determine Luke's house number

Mathematics
2 answers:
Olegator [25]3 years ago
7 0
I think it is b
----------------------------------------------
mash [69]3 years ago
7 0
First digit: A*1 =2 , A=2
Second digit: First digit*8 = 1*8 = 8
Third digit: 6-fourth digit => 6-6 =0 
Fourth digit: Fourth digit-First digit=5 => Fourth digit=5+first digit= 5+1 = 6

The house number= 1806.

The correct answer is A. 1806
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Find the points on the cone z2 = x2 + y2 that are closest to the point (6, 2, 0).
aliya0001 [1]

Answer:

The closest points on the cone are;

(6, 2, -√10) and (6, 2, √10)

Step-by-step explanation:

Let B(x, y, z) denote a point on the cone.

Therefore, the distance between the points (6, 2, 0) and B(x, y, z) is;

d = √[(x - 6)² + (y - 2)² + (z - 0)²]

d = √[(x - 6)² + (y - 2)² + z²]

Since we are given that z² = x² + y², we now have;

d = √[(x - 6)² + (y - 2)² + x² + y²]

Taking the square of both sides gives;

d² = [(x - 6)² + (y - 2)² + x² + y²]

x² is an increasing function. Thus, minimizing d is also the same as to minimize f (x, y) = d²

Thus, f' = 0. So;

df/dx = 2(x - 6) + 2x = 0

2x - 12 + 2x = 0

4x = 12

x = 12/4

x = 3

Similarly,

df/dy = 2(y - 2) + 2y = 0

2y - 4 + 2y = 0

4y - 4 = 0

4y = 4

y = 4/4

y = 1

Now,from earlier;

z² = x² + y²

Thus;

z = ±√(3² + 1²)

z = ±√10

Thus, the closest points on the cone are;

(6, 2, -√10) and (6, 2, √10)

7 0
3 years ago
If z1=12(cos⁡55°+isin⁡55°) and z2=4(cos⁡95°+isin⁡95°), then |z1z2|=
Y_Kistochka [10]

Answer:

This isn't middle school

Step-by-step explanation:

8 0
2 years ago
Math<br> Algebra 2!!!!!!!!!!!!!
tresset_1 [31]

Answer:

-1 < X ≤ 6

Step-by-step explanation:

x is greater than -1 but less than or equal to positive 6

8 0
3 years ago
Read 2 more answers
Find the measures of the numbered angles in each figure
inessss [21]
1 is 90o
2 is 90 - 27 = 63
3  63 angles 2 and 3 are in an isosceles triangle. They are opposite equal sides so the angles are equal.
6 0
3 years ago
Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disorde
zmey [24]

Answer:

t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638    

p_v =P(t_{(19)}>1.638)=0.0589    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

Step-by-step explanation:

Data given and notation    

\bar X=7.03 represent the sample mean

s=9.42 represent the sample standard deviation    

n=20 sample size    

\mu_o =3.58 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.    

z 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 mean is higher than 3.58 :    

Null hypothesis:\mu \leq 3.58    

Alternative hypothesis:\mu > 3.58    

Since we don't know the population deviation, 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{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638    

P-value    

First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

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

p_v =P(t_{(19)}>1.638)=0.0589    

Conclusion    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

7 0
3 years ago
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