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3 years ago

Enter the decimal 1.81 repeating as an improper fraction in simplest form.

Mathematics

1 answer:

Elodia [21]3 years ago

6 0

Answer:

1 9/11

Step-by-step explanation:

1.81 repeating is the equivalent to 1 9/11

AverageStudent

2 years ago

okay but like how?

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Will give brainliest!

Alecsey [184]

So when does y1 = y2?

at
2 (5x + 3) = (x2 -15)

solve and advise answer in comments and will assist further

7 0

3 years ago

Read 2 more answers

What is the reciprocal of 9/11

SVETLANKA909090 [29]

The reciprocal is the flipped value so it will be11/9

7 0

3 years ago

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Hey i need help with system of equations. i uploaded a doxs file you can onpen it in google docs

Art [367]

Answer:

Step-by-step explanation:

PART A:

The first equation is $x + y = 7\\y = 7 - x .............(a)$

The second equation is $3x - 2y = 1\\2y = 3x - 1$

Putting the value of y from (a) in the above equation, we get $2y = 3x - 1\\2(7 - x) = 3x - 1\\14 - 2x = 3x - 1\\x = 15$

From (a) we get y = 7 - x = 7 - 15 = -8

PART B:

Here the system of equations are $-x + 4y = 17........(a) and 3x - 5y = -30.........(b)$

From (a) we get, $x = 4y - 17$

Putting the above value of x in (b) we get, $3x - 5y = -30\\3(4y - 17) - 5y = -30\\12y - 51 - 5y = -30\\7y = 21\\y = 3$

Hence, $x = 4y - 17 = 12 - 17 = -5$

3 0

3 years ago

) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa

asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

Let $p_1$ = population proportion of drivers 16-24 years old who buckle up .

$p_2$ = population proportion of drivers 25-69 years old who buckle up .

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no significant difference in seat belt use}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a difference in seat belt use}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers 16-24 years old who buckle up = 79%

$\hat p_2$ = sample proportion of drivers 25-69 years old who buckle up = $\frac{924}{1100}$ = 84%

$n_1$ = sample of 16-24 years old drivers = 1000

$n_2$ = sample of 25-69 years old drivers = 1100

So, test statistics = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a difference in seat belt use.

4 0

3 years ago

Find "G". Round to two decimal places. G= R+ab/at R=257 a=4.88 b=37.4 t=61.5

givi [52]

Answer:

G ≈ 1.46

Step-by-step explanation:

Given

G = $\frac{R+ab}{at}$ , substitute values