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1 year ago

PLS HELP ASAP! GIVING BRAINLIST!

Mathematics

2 answers:

AleksAgata [21]1 year ago

7 0

Answer:

Step-by-step explanation:

luda_lava [24]1 year ago

7 0

Answer:

A. $10x^{2} + 33x^{2} +19x - 12$

Explanation:

$(2x + 3) (5^{2} + 9x - 4)\\$

Mutiply $2x$ by each number on the second bracket

$(2x) (5^{2} ) = 10x^{3} \\(2x) (9x ) = 18x^{2} \\(2x) (- 4 } ) = -8x$

Then mutiply 3 by each number on the second bracket

$(3) (5^{2} )= 15x^{2} \\(3) (9x) = 27x\\(3) (-4) = -12$

New equation:

$10x^{3} + 18x^{2} + -8x + 15^{2} + 27x - 12$

Add all like terms

$10x^{3} + 33x^{2} + 19x - 12$

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A recent survey shows that the probability of a college student drinking alcohol is 0.6. further, given that the student is over

Murrr4er [49]

Answer:

0.24

Step-by-step explanation:

These events are not mutually exclusive; this means they can happen at the same time.

For two events A and B that are not mutually exclusive,

P(A and B) = P(A) * P(B|A)

Let A be the event "over 21 years old" and B be the event "drinks alcohol".

The probability that a student is over 21 years old is 0.3; this is because 30% of the students are over 21 years old.

The probability that a student drinks alcohol given they are over 21 is 0.8.

This gives us

P(A and B) = 0.3(0.8) = 0.24

8 0

2 years ago

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Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

2 years ago

a model car is one tenth of the size of the real car the models measures 42cm longs what the length of length of the real car

trapecia [35]

Just multiply 42 by 10 and that gives you 420 , so the real car has a length of 420cm or 4.2 metres :)

8 0

1 year ago

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Solve y=3bx-7x for x

irina [24]

Y = 3bx - 7x
y = x(3b - 7)

Divide each side by 3b - 7 (assume that it is not zero).
$\frac{y}{3b-7} =x$

Answer:
$x= \frac{y}{3b-7}$

6 0

1 year ago

Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0=2x2+3x-8?

Basile [38]

Answer:

Step-by-step explanation:

The given quadratic equation is

2x^2+3x-8 = 0

To find the roots of the equation. We will apply the general formula for quadratic equations

x = -b ± √b^2 - 4ac]/2a

from the equation,

a = 2

b = 3

c = -8

It becomes

x = [- 3 ± √3^2 - 4(2 × -8)]/2×2

x = - 3 ± √9 - 4(- 16)]/2×2

x = [- 3 ± √9 + 64]/2×2

x = [- 3 ± √73]/4

x = [- 3 ± 8.544]/4

x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4

x = 5.544/4 or - 11.544/4

x = 1.386 or x = - 2.886

The positive solution is 1.39 rounded up to the nearest hundredth

8 0

2 years ago

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