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3 years ago

15

Please answer both I’ll give brainliest

1 answer:

Readme [11.4K]3 years ago

4 0

Answer:

1. 10x6=y

2. x-168=x

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There are 9 candidates total, 5 male and 4 female, who run for 4 committee positions. Suppose that each person is equally likely

gavmur [86]

Use the hypergeometric distribution.
M=number of Men=5
F=number of women=4
m=number of men elected=2
f=number ow women elected=2.

Assuming equal chance to get elected, then
P(2M,2F)=C(M,m)*C(F,f)/C(M+F,m+f)
=C(5,2)*C(4,2)/C(9,4)
=10*6/126
=10/21

Reference: Hypergeometric distribution.

6 0

3 years ago

Answer it plzzz plz <br>you would live long if correct answer

Ede4ka [16]

Answer: 3^15

Step-by-step explanation:

Explained in attached picture. Hope you'll find it easy to understand. If not, you can comment :)

8 0

2 years ago

Determine if the lines are parallel, perpendicular or neither y=-3/2x-4 and y=2/3x-7

aliya0001 [1]

Answer:

<u>They are perpendicular</u>

Step-by-step explanation:

Perpendicular lines have slopes that are negated versions of the other or positive versions of the slope of the other. The slopes are also flipped.

Parallel lines have the same slope.

Neither is when the lines do not follow either of these definitions.

The slope of your two lines are -3/2 and 2/3. These slopes are flipped and the opposite negations of each other. <u>This means the lines are perpendicular.</u>

<u />

4 0

3 years ago

How would you solve these if the X goes; -3,-2,-1,0,1,2,3 in a table format

Vlada [557]

Hello,

solution is jointed

7 0

3 years ago

When n=343 college students are randomly selected and surveyed, it is found that x=110 own a car. Find a 99% confidence interval

Liono4ka [1.6K]

Answer:

The margin of error will be "0.65". A further explanation is provided below.

Step-by-step explanation:

The given values are:

n = 343

x = 110

At 99% confidence level,

$\alpha = 1-99$ %

$=1-0.99$

$=0.01$

then,

$\frac{\alpha}{2} =\frac{0.01}{2}$

$=0.005$

or,

$Z_{\frac{\alpha}{2} }=Z_{0.005}$

$=2.576$

Now,

The point estimate will be:

⇒ $\hat{P}=\frac{x}{n}$

⇒ $=\frac{110}{343}$

⇒ $=0.321$

or,

⇒ $1-\hat{P}=1-0.321$

⇒ $=0.679$

The margin of error will be:

⇒ $E=Z_{\frac{\alpha}{2} }\times \sqrt (\frac{(\hat{P}\times (1 - \hat{P})) }{n} )$

On substituting the above values, we get

⇒ $=2.576\times \sqrt{\frac{0.321\times 0.679}{343} }$

⇒ $=2.576\times \sqrt{\frac{0.217959}{343} }$

⇒ $=0.065$

hence,

⇒ $\hat{P}-E$

⇒ $0.321-0.065$

⇒ $0.256,0.386$

6 0

3 years ago