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Black_prince [1.1K]
3 years ago
11

If 5 + 20 times 2 Superscript 2 minus 3 x Baseline = 10 times 2 Superscript negative 2 x Baseline + 5, what is the value of x? –

3 –2 2 3
Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0

Answer:

d

Step-by-step explanation:

Just did it.

sp2606 [1]3 years ago
5 0

Answer:

<h3>Option D) 3 is correct</h3><h3>Therefore the value of x is 3</h3>

Step-by-step explanation:

Given equation is 5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5

<h3>To find the value of x :</h3>

First solving the given equation we have,

5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5

5+20\times (2)^{2-3x}-5=10\times (2)^{-2x}+5-5

20\times (2)^{2-3x}=10\times (2)^{-2x}

20\times (2)^2.(2)^{-3x}=10\times (2)^{-2x}

20\times 4.(2)^{-3x}=10\times (2)^{-2x}

80(2)^{-3x}=10\times (2)^{-2x}

\frac{80}{10}(2)^{-3x}=\frac{10\times (2)^{-2x}}{10}

8(2)^{-3x}=(2)^{-2x}

\frac{(2)^{-3x}}{(2)^{-2x}}=\frac{1}{8}

(2)^{-3x}.(2)^{2x}=\frac{1}{2^3} ( by using the property \frac{1}{a^{-m}}=a^m )

2^{-3x+2x}=\frac{1}{2^3} ( by using the property a^m.a^n=a^{m+n} )

2^{-x}=\frac{1}{2^3}

\frac{1}{2^x}=\frac{1}{2^3} ( by using the property a^{-m}=\frac{1}{a^m} )

Since bases are same so powers are same

Therefore we can equate the powers we get x=3

<h3>Therefore the value of x is 3</h3><h3>Option D) 3 is correct</h3>
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For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

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maks197457 [2]
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5) A submarine is cruising at -40 meters (40 meters below the surface). It
Verdich [7]

Answer:

-25

Step-by-step explanation:

You begin with -40, then decrease by 20 so your new value is -60, then increase by 35 so your final value is -25m... hope this helps (:

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Given that y is directly proportional to x^2 and that y=2 when x=3 find the value of y when x=6
Sergio039 [100]

Answer:y=4/3

y is directly proportional to x^2 is written as

y \alpha  {x}^{2} introducing a constant,

y=kx^2

but from the question, when y=2 , x=3 . putting it in the formula to get the value of k

2 =  {3}^{2}  \times k

2=9k . <em>divi</em><em>ding</em><em> </em><em>throu</em><em>gh</em><em> </em><em>by</em><em> </em><em>9</em><em> </em><em>to</em><em> </em><em>get</em><em> </em><em>the</em><em> </em><em>valu</em><em>e</em><em> </em><em>of</em><em> </em><em>k</em>

<em>\frac{2}{9}  =  \frac{9k}{9}</em>

<em>k =  \frac{2}{9}</em>

<em>pu</em><em>tting</em><em> </em><em>it</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>ge</em><em>neral</em><em> </em><em>expres</em><em>sion</em>

<em>y =  \frac{2}{9} x</em>

<em>the</em><em> </em><em>valu</em><em>e</em><em> </em><em>of</em><em> </em><em>y</em><em> </em><em>whe</em><em>n</em><em> </em><em>x</em><em>=</em><em>6</em>

<em>y =  \frac{2}{9}  \times 6</em>

<em>y =  \frac{4}{3}</em>

<em>there</em><em>fore</em><em> </em><em>the</em><em> </em><em>valu</em><em>e</em><em> </em><em>of</em><em> </em><em>y</em><em> </em><em>when</em><em> </em><em>x</em><em>=</em><em>6</em><em> </em><em>is</em>

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3 0
3 years ago
A recent estimate by a large distributor of gasoline claims that 60% of all cars stopping at their service stations chose unlead
Anton [14]

Answer:

We reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

Step-by-step explanation:

There are 3 types of gas listed in the question.

Thus;

n = 3

DF = n - 1

DF = 3 - 1

DF = 2

Let's state the hypotheses;

Null hypothesis; H0: P_regular = P_super unleaded = 20%; P_i leaded = 60%

Alternative hypothesis; Ha: At least 2 proportions differ from the stated value.

Observed values are;

Regular gas; O = 51

Unleaded gas; O = 261

Super Unleaded; O = 88

Total observed values = 51 + 261 + 88 = 400

We are told that super unleaded and regular were each selected 20% of the time and that unleaded gas was chosen 60% of the time.

Thus, expected values are;

Regular gas; E = 20% × 400 = 80

Unleaded gas; E = 60% × 400 = 240

Super Unleaded; E = 20% × 400 = 80

Formula for chi Square goodness of fit is;

X² = Σ[(O - E)²/E]

X² = (51 - 80)²/80) + (261 - 240)²/240) + (88 - 80)²/80)

X² = 13.15

From the chi Square distribution table attached and using; DF = 2 and X² = 13.15, we can trace the p-value to be approximately 0.001

Also from online p-value from chi Square calculator attached, we have p to be approximately 0.001 which is similar to what we got from the table.

Now, if we take the significance level to be 0.05, it means the p-value is less than it and thus we reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

4 0
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