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3 years ago

chase is using 4 bags of birdseed to fill 12 bird feeders. What part of a bag will be in each bird feeder

Mathematics

1 answer:

tiny-mole [99]3 years ago

6 0

Each bird feeder will contain 1/3 a bag of bird seed

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Last month you had 8 wins while playing fortnight. This month you had 5. What was your percent of change?

Dominik [7]

Answer:

%62.5

Step-by-step explanation:

Im pretty sure you jsut divide 5/8 and then multiply by 100 to get a percent

5/8=.625

.625*100=62.5

%62.5

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3 years ago

Undecillion how many 0s.

Irina-Kira [14]

Answer:

<u>Undecillion</u> has 36 0's

$\sf{\#FromThePhilippines}$

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2 years ago

Help fast will give brainlist

Scrat [10]

Answer:

the picture is black send another one

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3 years ago

X is a normally Distributed random variable with a standard deviation of 4.00. Find the mean of X when 64.8% of the area lies to

Cerrena [4.2K]

Answer:

Step-by-step explanation:

σ = 4 ; μ =?

8.52 to the left of X

P(X < 8.52) = 64.8%

P(X < 8.52) = 0.648

Using the Z relation :

(x - μ) / σ

P(Z < (8.52 - μ) / 4)) = 0.648

The Z value of 0.648 of the lower tail is equal to 0.38 (Z probability calculator)

Z = 8.52 - μ / 4

0.38 = 8.52 - μ / 4

0.38 * 4 = 8.52 - μ

1.52 = 8.52 - μ

μ = 8.52 - 1.52

μ = 7

5 0

3 years ago

Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2.

insens350 [35]

Answer:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

Step-by-step explanation:

Given

$p(x) = \frac{x^2-2x-3}{x+2}$ -- Missing from the question

Required

The behavior of the function around its vertical asymptote at $x = -2$

$p(x) = \frac{x^2-2x-3}{x+2}$

Expand the numerator

$p(x) = \frac{x^2 + x -3x - 3}{x+2}$

Factorize

$p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}$

Factor out x + 1

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

$x -> -2^{-}$ Say x = -3

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

$p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12$

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

$x->-2^{-}, p(x)->-\infty$

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: $x>-2$

Say x = -2.1

$p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1$

We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

$x->-2^{+}, p(x)->-\infty$

So, the behavior is:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

6 0

3 years ago