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2 years ago

Find the value of x, rounded to the nearest tenth.

Mathematics

1 answer:

Brrunno [24]2 years ago

7 0

Step-by-step explanation:

first find the height by using Tan 28
then use pythagras theorem you can find value of x
answer is 36.2

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An experiment consists of rolling a die, flipping a coin, and spinning a spinner divided into 7 equal regions. The number of ele

oee [108]

Answer:

Option A.

The number of elements in the sample space of this experiment is 84

Step-by-step explanation:

The sample space for an experiment represents the number of possible results that may occur when conducting the experiment.

When throwing a die and observe the number obtained there are 6 possible results

When throwing a coin there are 2 possible results.

When spinning a wheel divided into 7 regions we have 7 possible outcomes.

If we combine these three experiments, then the sample space S is composed of the product of the possible results of each experiment.

$S = (6)(2)(7)\\\\S =84$

5 0

3 years ago

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

2 years ago

Classify each of the power functions based on their end behavior (increasing or decreasing) as x = ∞

Svetlanka [38]

Answer: right side behavior:

f(x) is Decreasing

g(x) is Increasing

h(x) is Increasing

j(x) is Decreasing

Step-by-step explanation:

The rules for end behavior are based on 2 criteria: Sign of leading coefficient and Degree of polynomial

Sign of leading coefficient (term with greatest exponent):

If sign is positive, then right side is increasing
If sign is negative, then right side is decreasing

Degree of polynomial (greatest exponent of polynomial:

If even, then end behavior is the same from the left and right
If odd, then end behavior is opposite from the left and right

f(x) = -2x²

Sign is negative so right side is decreasing
Degree is even so left side is the same as the right side (decreasing)

as x → +∞, f(x) → +∞ Decreasing

as x → -∞, f(x) → -∞ Decreasing

g(x) = (x + 2)³

Sign is positive so right side is increasing
Degree is odd so left side is opposite of the right side (decreasing)

as x → +∞, f(x) → +∞ Increasing

as x → -∞, f(x) → -∞ Decreasing

$h(x)=-1+x^{\frac{1}{2}}\implies h(x)=x^{\frac{1}{2}}-1$

Sign is positive so right side is increasing
Degree is an even fraction so left side is opposite of the right side as it approaches the y-intercept (-1)

as x → +∞, f(x) → +∞ Increasing

as x → -∞, f(x) → -1 Decreasing to -1

$j(x)=\dfrac{1}{2}(-x)^5\implies j(x)=\dfrac{1}{2}(-1)^5(x)^5\implies j(x)=-\dfrac{1}{2}x^5$