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Snezhnost [94]
3 years ago
13

When x is 2, y is 4, p is 0.5, and m is 2. If x varies directly with the product of p and m and inversely with y, which equation

models the situation?
Mathematics
2 answers:
Nesterboy [21]3 years ago
8 0

Answer:

x=8((m*p)/y)

Step-by-step explanation:

You have that x varies directly with p*m (which means that if the result of m*p increase, x increase)

And you have too that x is inversely with y (which means that if y increase, x decrease).

Then you must get:

x=constant value((m*p)/y)

If the problem says that x is 2, y is 4, p is 0.5, and m is 2

the constant value should be 8, in this way:

x=8((m*p)/y)

If you replace:

x=8((2*0,5)/4)

The result is x=2

galben [10]3 years ago
3 0
Can you elaborate please? What does p and m mean?
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Solve the inequality and graph the solution on the line provided.<br> -7x + 8&lt; -6
HACTEHA [7]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

-7x+8<-6  Equation/Question

Step 2

-7x+8<-6  Simplify

-7x+8<-6

Step 3

-7x+8<-6  Subtract 8

-7x<-14

Step 4

-7x<-14  Divide sides by -7

answer

x>2

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7 0
3 years ago
A hockey team is convinced that the coin used to determine the order of play is weighted. The team captain steals this special c
fredd [130]

Answer:

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

Step-by-step explanation:

Let p be the probability of heads in a single toss of the coin. Then our null hypothesis that the coin is fair will be formulated as

H0 :p 0.5   against   Ha: p ≠ 0.5

The significance level is approximately 0.05

The test statistic to be used is number of heads x.

Critical Region: First we compute the probabilities associated with X the number of heads using the binomial distribution

Heads (x)        Probability (X=x)                        Cumulative     Decumulative

0                        1/16384 (1)             0.000061     0.000061

1                         1/16384  (14)         0.00085             0.000911

2                       1/16384 (91)           0.00555             0.006461

3                       1/16384(364)         0.02222

4                       1/16384(1001)         0.0611

5                       1/16384(2002)       0.122188

6                        1/16384(3003)      0.1833

7                         1/16384(3432)      0.2095

8                        1/16384(3003)       0.1833

9                        1/16384(2002)       0.122188

10                       1/16384(1001)        0.0611

11                       1/16384(364)        0.02222

12                      1/16384(91)            0.00555                             0.006461

13                     1/16384(14)              0.00085                           0.000911

14                       1/16384(1)            0.000061                            0.000061

We use the cumulative and decumulative column as the critical region is composed of two portions of area ( probability) one in each tail of the distribution. If  alpha = 0.05 then alpha by 2 - 0.025 ( area in each tail).

We observe that P (X≤2) =   0.006461 > 0.025

and

P ( X≥12 ) = 0.006461 > 0.025

Therefore true significance level is

∝=  P (X≤0)+P ( X≥14 ) = 0.000061+0.000061= 0.000122

Hence critical region is (X≤0) and ( X≥14)

Computation x= 12

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

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