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3 years ago

I Give Brainiest

Mathematics

1 answer:

iragen [17]3 years ago

5 0

Answer:

X-42.7

Y-5.34

Step-by-step explanation:

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If f(x) = -5-4 and g(x) = -3 -2 find (f-g)(x)

asambeis [7]

<u>Answer:</u>

The correct answer option is D. $-5^x+3x-2$

<u>Step-by-step-explanation:</u>

We are given the following two functions:

$f(x)=-5^x-4$

and

$g(x)= -3x-2$

We are to find $(f-g)(x)$ so we will subtract the function g from function f, arrange the like terms together and combine them like shown below to get:

$(f-g)(x) =(-5^x-4)-(-3x-2)$

$(f-g)(x)=-5^x-4+3x+2$

$(f-g)(x)=-5^x+3x-2$

Therefore, $(f-g)(x)$ is equal to $-5^x+3x-2$ so the correct answer option is D. $-5^x+3x-2$ .

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3 years ago

Which equation represents this statement?

yanalaym [24]

Answer:

first one

hope it helps you ...

plzzz plzzz plzzz mark as brainliest

7 0

2 years ago

............... help me please i don’t understand this

Sergio [31]

Answer:

x = 11

Step-by-step explanation:

3 0

3 years ago

Tyler bought a large bag of peanuts at a baseball game. Is it more reasonable to say that the mass of the peanuts is 1 gram or 1

Blababa [14]

Would you want only 1 gram of peanuts?
Or one Pound of peanuts? or kilograms

1 Kilo

4 0

3 years ago

Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly sel

laiz [17]

Answer:

0.03 is the probability that for the sample mean IQ score is greater than 103.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6$

P( mean IQ score is greater than 103)

P(x > 103)

$P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)$

$= 1 - P(z \leq 1.875)$

Calculation the value from standard normal z table, we have,

$P(x > 103) = 1 - 0.970 =0.03= 3\%$

0.03 is the probability that for the sample mean IQ score is greater than 103.

7 0

3 years ago