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3 years ago

Estion 3/10

Mathematics

2 answers:

Rus_ich [418]3 years ago

4 0

Answer:

I believe its c but I'm not clearly sure

AlladinOne [14]3 years ago

3 0

Answer:

i think the answer is d if im not u can report me

Step-by-step explanation:

plz mark as brainliest

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Solve the system with elimination. x - y = 10 3x - 2y = 25

raketka [301]

Answer:

Step-by-step explanation:

{x,y}={5,-5}

8 0

2 years ago

The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5

Lena [83]

So the mean is 72.97

We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333

The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8

7 0

3 years ago

The gym teacher has $250 to spend on volleyball equipment. She buys 4 volleyball nets for $28 each. Volleyballs cost $7 each. Ho

pav-90 [236]

Answer:

The maximum number of volleyballs that she can buy is 19

Step-by-step explanation:

Let

x ----> the number of volleyballs

we know that

The cost of each volleyball net ($28) by the number of volleyball nets (4) plus the cost of each volleyball ($7) multiplied by the number of volleyballs (x) must be less than or equal to $250

The inequality that represent this situation is

$28(4)+7x\leq 250$

Solve for x

$112+7x\leq 250$

subtract 112 both sides

$7x\leq 250-112$

$7x\leq 138$

Divide by 7 both sides

$x\leq 19.7$

therefore

The maximum number of volleyballs that she can buy is 19

4 0

3 years ago

The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100

Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, $\mu$ = 310/1000 or 0.31 and Standard deviation, $\sigma$ = 14.63/1000 = 0.01463 .

We know that Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

P(X <=0.13) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{0.13-0.31}{0.01463}$ ) = P(Z <= -12.303) = P(Z > 12.303)

Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.

So, after seeing the table we can say that this probability is approximately less than 0.0005% .

4 0

3 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

3 years ago