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Nana76 [90]
3 years ago
9

Look at this table: x 19 5 19 y 2 5 8 Is this relation a function? yes no​

Mathematics
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

no

Step-by-step explanation:

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the answer is B)13800

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Please help, solve for the value of each variable.
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4 0
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Solve -8x + 4 &lt; 36.<br><br>A. x &gt; 4<br>B. x &lt; 4<br>C. x &lt; -4<br>D. x &gt; -4​
andre [41]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

let's solve for x ~

  • - 8x + 4 < 36

  • - 8x < 36 - 4

  • - 8x < 32

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6 0
3 years ago
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-6x + 18 = 7- (4x + 9) help will give brainlyist
andreyandreev [35.5K]

Answer:

X = -5

Step-by-step explanation:

-6x + 18 = 7 - (4x + 9)

first we simplify the parentheses.

distribute the invisible number one created by the negative sign in front of the parentheses to get -6x + 18 = 7 - 4x - 9

then simplify further by subtracting 9 from the right side, leaving you with

-6x + 18 = -2 - 4x

next, add 2 to both sides to isolate the variable.

-6x + 20 = -4x

then, to isolate the variable further add 6x to both sides.

20 = -4x

divide both sides by -4 to isolate the variable

-5 = x

so, x = -5

6 0
3 years ago
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According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

5 0
4 years ago
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