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2 years ago

What is the value of the expression when n = 3? - 2n(5 + n-8-3n)

Mathematics

1 answer:

lys-0071 [83]2 years ago

5 0

Answer:

54

Step-by-step explanation:

- 2n(5 + n - 8 - 3n) =

replace n = 3

- 2(3)(5 + 3 - 8 - 3×3)

= - 6(8 - 8 - 9)

= - 6(0 - 9)

= - 6 × - 9

= 54

(-)(-) = +

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ratelena [41]

Answer:

Step-by-step explanation:

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2 years ago

in the figure p is the incenter of isosceles rst what type of triangle is rpt? Thank you in advance .

snow_lady [41]

Answer:

$\Delta \text{RPT is an isosceles triangle}$

Step-by-step explanation:

As in the triangle RST,

$RS=ST\ \ \ \Rightarrow m\angle SRT=m\angle STR$

Because if in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

Incenter is obtained by the Intersection of a triangle's three angle bisectors and it is equidistant from the three sides of the triangle.

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2 years ago

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The amount of time required to reach a customer service representative has a huge impact on customer satisfaction. Below is the

Vika [28.1K]

Answer:

The value of the test statistic is t=1.12.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the mean amount of time required to reach a customer service representative significantly differs between the two hotels.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The sample 1, of size n1=20 has a mean of 2.65 and a standard deviation of √2.952=1.72.

The sample 2, of size n2=20 has a mean of 2.01 and a standard deviation of √2.952=1.89.

The difference between sample means is Md=0.64.

$M_d=M_1-M_2=2.65-2.01=0.64$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{1.72^2+1.89^2}{20}}\\\\\\s_{M_d}=\sqrt{\dfrac{6.531}{20}}=\sqrt{0.327}=0.571$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.64-0}{0.571}=\dfrac{0.64}{0.571}=1.12$

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2 years ago

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2 years ago

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let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo

anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

$f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\$

a) P(x > 5) = $\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865$

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = $\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}$

But P(x > a) = 0.05

Therefore:

$e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98$

a ≅ 12

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2 years ago