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2 years ago

PLS HELP!! DUE TONIGHT!! 15 POINTS!!

Mathematics

2 answers:

aliya0001 [1]2 years ago

7 0

Answer:

the answer for this is c

denpristay [2]2 years ago

3 0

Answer:

The answer is C

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Help me with this! Solve for x

olganol [36]

Answer:

x=6

Step-by-step explanation:

Line ZW and YX are parallel, which means they have the same measurements.

Therefore:

3x+2=20

subtract two from both sides

3x=18

divide by 3

3x/3=18/3

x=6

hope this helps =D

5 0

3 years ago

Triangle QRS is similar to triangle TUV. Using the image below, prove that lines RS and UV have the same slope. You must show al

N76 [4]

Explanation:

Lines will have same slope if they make the same angle with respect to the vertical or horizontal

The similarity statement ∆QRS ~ ∆TUV tells you that ∠R ≅ ∠U.

Angle R is the angle measured clockwise from vertical segment RQ to segment RS. Angle U is the angle measured clockwise from vertical segment UT to segment UV.

Segments RS and UV both make the same angle with a vertical segment, so have the same slope.

6 0

2 years ago

How many fifths are there in 3⅗

Ilya [14]

Answer:

18/5

Step-by-step explanation:

5 1/5 = 1 thus 3 are 15 + 3 = 18/5

8 0

3 years ago

Read 2 more answers

Which statement describes the end behavior of the exponential function f(x) = 2^x-3?

Travka [436]

Lim 2^x-3 = lim 2^(1/x^3) = 2^(1/inf) = 2^0 = 1

x -> infinity

The function approaches 1

4 0

4 years ago

A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago