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4 years ago

How to solve for y if the equation is y=5/2x + 5 if x is zero

2 answers:

6 0

Q, y=5/2x+x if x is zero

An, ( y=5/2×0+5 ) as you know if i multiply number to zero you get zero so "x" change to the zero you get
( y=0+5 )= ( y=5 ) that is your answer

Karolina [17]4 years ago

5 0

$y=\dfrac{5}{2}\cdot0+5\\
y=0+5\\
y=5$

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Rachel and her cousin Lena live in different cities about 100 miles apart.On the map the two cities measure 5 inches apart . Wha

Ostrovityanka [42]

On her map 1 inch is equal to 20 miles.

5x=100

divide both sides by 5

x=20

5 is for the 5 inches.
x is for the unknown.
100 is the 100 miles between the cities.

So the scale 1:20

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4 years ago

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3 years ago

Find each product (4x - 5)(2x5 + x3 – 1)

TEA [102]

Answer:

8x^6-10x^5+4x^4-5x^3-4x+5

Step-by-step explanation:

Lest distribute the expression given

this is multiply the first term (4x) with the longer expression (2x^5+x^3-1) and then, multiply the second term (-5) with the longer expression (2x^5+x^3-1).

When the above is finish, simply sum all together.

4x* (2x^5+x^3-1)= 8x^6 +4x^4-4x

-5*(2x^5+x^3-1)=-10x^5-5x^3+5

The sum of all together (respecting the coefficient they belong) is shown:

8x^6-10x^5+4x^4-5x^3-4x+5

8 0

4 years ago

2. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of

IgorC [24]

Answer:

a) 37.31 b) 42.70 c) 0.57 d) 0.09

Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., $\mu$ = 35 and $\sigma$ = 6. We know that the probability density function for a normal distribution with a mean of $\mu$ and a standard deviation of $\sigma$ is given by

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]$

in this case we have

$f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]$

Let $X$ be the random variable that represents a row score, we find the values we are seeking in the following way

a) we are looking for a number $x_{0}$ such that

$P(X\leq x_{0})$ = $\int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65$ , this number is $x_{0}$ =37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number $x_{1}$ such that

$P(X\leq x_{1})$ = $\int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9$ , this number is $x_{1}$ =42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

$P(28\leq X\leq 38)$ = $\int\limits^{38}_{28} {f(x)} \, dx = 0.57$

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

$P(41\leq X\leq 44)$ = $\int\limits^{44}_{41} {f(x)} \, dx = 0.09$

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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4 years ago

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padilas [110]

Option C is the correct answer.

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3 years ago