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3 years ago

What's the lim (x^5-x^3+2)/ (®(x+3)-2)

Mathematics

1 answer:

igomit [66]3 years ago

6 0

Answer:

send here I answer

Step-by-step explanation:

za r i misa 0 g mail

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como calcular la cantidad de horas por proyecto si un contratista tiene 5,000 horas-hombres y el costo es de $10 y $12 por proye

Tju [1.3M]

Answer:

Each project has 2.500 man-hours

Step-by-step explanation:

Available man hours: 5000

For two projects Costs $

X: project A 10

Y: project B 12

Approach:

X + Y = 5000 => X = 5000 - Y

10 * X + 12 * Y = 55000

replacing:

10 * (5000 - Y) + 12 * Y = 55000

50000 - 10 * Y + 12 * Y = 55000

2 * Y = 55000 - 50000

Y = 5000/2

Y = 2500

For X:

X = 5000 - 2500

X = 2500

The number of hours available for each project are:

Projects Man hours

X: project A 2500

Y: project B 2500

6 0

3 years ago

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

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

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.96 \cdot 475.795=932.54$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13$

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

6 0

3 years ago

Luisa recieved $108 for a necklace she sold on an online auction. The auction site took a fe of 10% of selling price. What was t

Vera_Pavlovna [14]

I believe it might be 107.9 after if that’s what your asking-like taking off 10% of $108.

3 0

3 years ago

How do you solve 90% of 20?

Marizza181 [45]

All you would do is multiply .9 by 20
.9*20=18

90% of 20 is 18

3 0

3 years ago

A group of 3 friends spends a total of $17.25 for brunch at a restaurant before tax and tip. Everyone orders the same thing: scr

liubo4ka [24]

Answer:

17.25/3 = 5.75

Step-by-step explanation:

You divide 17.25 by 3.

It equals 5.75. Each friend has 5.75 on spend on brunch.

Toast is $1. Orange Juice is $1.75. Eggs are $2.

Hope this helps.

3 0

3 years ago

What's the lim (x^5-x^3+2)/ (®(x+3)-2)​

What's the lim (x^5-x^3+2)/ (®(x+3)-2)