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Virty [35]
3 years ago
8

Math question #1 please show steps

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

C

Step-by-step explanation:

An approximation of an integral is given by:

\displaystyle \int_a^bf(x)\, dx\approx \sum_{k=1}^nf(x_k)\Delta x\text{ where } \Delta x=\frac{b-a}{n}

First, find Δx. Our a = 2 and b = 8:

\displaystyle \Delta x=\frac{8-2}{n}=\frac{6}{n}

The left endpoint is modeled with:

x_k=a+\Delta x(k-1)

And the right endpoint is modeled with:

x_k=a+\Delta xk

Since we are using a Left Riemann Sum, we will use the first equation.

Our function is:

f(x)=\cos(x^2)

Therefore:

f(x_k)=\cos((a+\Delta x(k-1))^2)

By substitution:

\displaystyle f(x_k)=\cos((2+\frac{6}{n}(k-1))^2)

Putting it all together:

\displaystyle \int_2^8\cos(x^2)\, dx\approx \sum_{k=1}^{n}\Big(\cos((2+\frac{6}{n}(k-1))^2)\Big)\frac{6}{n}

Thus, our answer is C.

*Note: Not sure why they placed the exponent outside the cosine. Perhaps it was a typo. But C will most likely be the correct answer regardless.

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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
Samira needs 22 quarts of lemonade to fill a dispenser. She plans to buy 7 jugs of lemonade. Each jug contains 3.3 quarts.
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I think its b iam not to sure i have the same question myself
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3 years ago
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Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam
frosja888 [35]

Answer:

Advance tickets cost $30; same-day tickets cost $35.

Step-by-step explanation:

Let a = the cost of an advance ticket

and s = the cost of a same-day ticket

We have two conditions:

                                                         (1)                  a + s = 65

                                                         (2)        15a + 20s = 1150

Subtract a from each side of  (1)     (3)                       s = 65 - a

Substitute (3) into (2)                         15a + 20(65 - a) = 1150  

Distribute the 20                             15a + 1300 - 20a  = 1150

Combine like terms                                     1300 - 5a = 1150

Subtract 1300 from each side                               -5a = -150

Divide each side by -5                    (4)                     a = 30

Substitute (4) into (1)                                          30 + s = 65

Subtract 30 from each side                                     s = 35

Advance tickets cost $30; same-day tickets cost $35.

Check:

(1) 30 + 35 = 65     (2) 15 × 30 + 20 × 35 = 1150

            65 = 65             450    +    700    = 1150

                                                         1150 = 1150

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What is the standard form of 46
Korolek [52]

Answer:

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Step-by-step explanation:

46 is already in standard form

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Which triangle could be drawn as it is described?
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Step-by-step explanation:

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