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3 years ago

ZDAC

Mathematics

1 answer:

n200080 [17]3 years ago

7 0

Answer:

7.3

Step-by-step explanation:

right for me on Khan

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What is the equation of the line in slope-intercept form?

lutik1710 [3]

The slope is negative and is -16 / 8 = -2

The y intercept is at y = 16

General form is y = mx + c
m = -2 and c = 16 so we have :-

y = -2x + 16 Answer

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4 years ago

Helppppp plz 10points

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3 years ago

Ebony earned scores of 72 70, 81, 99, and 88 on five math tests. Determine her

Minchanka [31]

Let it be x

$\boxed{\sf Average=\dfrac{Sum\:of\:terms}{No\:of\:terms}}$

$\\ \sf\longmapsto \dfrac{72+70+81+99+88+x}{6}=83$

$\\ \sf\longmapsto \dfrac{410+x}{6}=83$

$\\ \sf\longmapsto 410+x=498$

$\\ \sf\longmapsto x=498-410$

$\\ \sf\longmapsto x=88$

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3 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion 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 males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

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

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

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

95% confidence interval for ( $p_1-p_2$ ) =

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

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

Which numbers have a digit in the ones place that is 1/10 the value of the digit in the tens place

liq [111]

Answer:

4,099 and 5,011

Step-by-step explanation:

This problem can be solved by taking options one by one.

Option (1) : 4,099

Digit in ones place = 9

The value of the digit in tens place = 90

$\dfrac{9}{90}=\dfrac{1}{10}$ . It is correct.

Option (2) : 4,110

Digit in one places = 0

The value of the digit in tens place = 10

It is incorrect.

Option (3) : 5,909

Digit in one places = 9

The value of the digit in tens place = 0

It is again incorrect.

Option (4) : 5,011

Digit in one places = 1

The value of the digit in tens place = 10

$\dfrac{1}{10}$ . It is correct.

Hence, in option (a) and (d), the he ones place is 1/10 the value of the digit in the tens place.

7 0

4 years ago