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3 years ago

5

Bill and Dave collect baseball cards. They have 111 cards in total. Dave has twice as Bill. how many cards does each of them hav

e.

1 answer:

never [62]3 years ago

3 0

Answer:Dave has 74 and Bill has 37

Explanation:111÷3=37×2=<u>74</u>

111-74=<em>37</em>

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5k/2G is the answer to the problem.

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A number is being written in scientific notation. The student moves the decimal point to the left 3 places and _____.

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There is only one negative exponent here, and it is the first option.
So:
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3 years ago

Read 2 more answers

A rectangle is inscribed in a right isosceles triangle, such that two of its vertices lie on the hypotenuse, and two other on th

Rufina [12.5K]

Answer:

Part 1) The base is $25\ in$ and the height is $10\ in$

Part 2) The base is $7.5\ in$ and the height is $18.75\ in$

Step-by-step explanation:

case 1) Right isosceles triangle of the left

Let

x------> the base of the rectangle

y----> the height of the rectangle

Remember that

In a right isosceles triangle the lengths of the legs of the triangle is the same

$y+x+y=45$

$2y+x=45$ ----> equation A

$\frac{x}{y} =\frac{5}{2}$

$x=2.5y$ -----> equation B

substitute equation B in the equation A

$2y+2.5y=45$

$4.5y=45$

$y=10\ in$

Find the value of x

$x=2.5(10)=25\ in$

case 2) Right isosceles triangle of the right

Let

x------> the base of the rectangle

y----> the height of the rectangle

Remember that

In a right isosceles triangle the lengths of the legs of the triangle is the same

$y+x+y=45$

$2y+x=45$ ----> equation A

$\frac{y}{x} =\frac{5}{2}$

$y=2.5x$ -----> equation B

substitute equation B in the equation A

$2(2.5x)+x=45$

$6x=45$

$x=7.5\ in$

Find the value of y

$y=2.5(7.5)=18.75\ in$

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3 years ago

What are the common factors that Greg found?

Arada [10]

<span>The answer is: 1, 2, 7, 14 .
</span>_____________________________________________________
Explanation:
________________________________________________
FIrst, list all the factors of 42 (positive integers):
________________________________________________________
1, 42
2, 21,
3, 14,
6, 7
_________________________________________________
Then list all the factors of 28 (positive integers):
____________________________________________________
<span> 1, 28
2, 14,
4, 7
</span>_____________________________________________________
Now, list all numbers that occur for BOTH the factors of "28" AND the factors of "42"
_____________________________________________________
1, 2, 7, 14
_____________________________________________________
The answer is: 1, 2, 7, 14 .
_______________________________________________________

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3 years ago

hilip Morris wishes to determine if there is a difference between the proportion of women and proportion of men who smoke cigare

murzikaleks [220]

Answer:

We conclude that the proportion of women who smoke cigarettes is smaller than or equal to the proportion of men at 0.01 significance level.

95% confidence interval for the difference in population proportions of women and men who smoke cigarettes is [0.0062 , 0.1298].

Step-by-step explanation:

We are given that random samples of 125 women and 140 men reveal that 13 women and 5 men smoke cigarettes.

<em>Let </em> $p_1$ <em> = population proportion of women who smoke cigarettes</em>

<em /> $p_2$ <em> = population proportion of men who smoke cigarettes</em>

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$ {means that the proportion of women who smoke cigarettes is smaller than or equal to the proportion of men}

Alternate Hypothesis, $H_A$ : $p_1> p_2$ {means that the proportion of women who smoke cigarettes is higher than the proportion of men}

The test statistics that will be used here is <u>Two-sample z proportion test</u> <u>statistics</u>;

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 women who smoke cigarettes= $\frac{13}{125}$ =0.104

$\hat p_2$ = sample proportion of men who smoke cigarettes = $\frac{5}{140}$ = 0.036

$n_1$ = sample of women = 125

$n_2$ = sample of men = 140

So, <u><em>the test statistics</em></u> = $\frac{(0.104-0.036)-(0)}{\sqrt{\frac{0.104(1-0.104)}{125}+ \frac{0.036(1-0.036)}{140}} }$

= 2.158

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is less than the critical value of z as 2.158 < 2.3263, so we have insufficient evidence to reject our null hypothesis due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the proportion of women who smoke cigarettes is smaller than or equal to the proportion of men.

<em>Now, coming to 95% confidence interval;</em>

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportions is given by;

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

where, $\hat p_1$ = sample proportion of women who smoke cigarettes= $\frac{13}{125}$ =0.104

$\hat p_2$ = sample proportion of men who smoke cigarettes = $\frac{5}{140}$ = 0.036

$n_1$ = sample of women = 125

$n_2$ = sample of men = 140

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between population proportions, </u> $(p_1-p_2)}$ <u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < $(p_1-p_2)}$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

<u>95% confidence interval for</u> $(p_1-p_2)}$ =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.104-0.036)-1.96 \times {\sqrt{\frac{0.104(1-0.104)}{125}+ \frac{0.036(1-0.036)}{140}} }$ , $(0.104-0.036)+1.96 \times {\sqrt{\frac{0.104(1-0.104)}{125}+ \frac{0.036(1-0.036)}{140}} }$ ]

= [0.0062 , 0.1298]

Therefore, 95% confidence interval for the difference in population proportions of women and men who smoke cigarettes is [0.0062 , 0.1298].

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3 years ago