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4 years ago

If x = 6 and y= 4, evaluate the following expression: 3x^2 + 3xy + y^2

Mathematics

2 answers:

Shkiper50 [21]4 years ago

4 0

Answer:

plz refer the attachment

EleoNora [17]4 years ago

4 0

Answer:

3*6^2 + 3*6*4 + 4^2

18^2 + 72 + 4^2

324 + 72 + 16

412

Answer: 412

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What is the answer for the second one?<br> 7% is the same as ??

ipn [44]

Answer:

0.07

Step-by-step explanation:

Just put the value of the percent

40, 90, or 7 over 100

40/100, 90/100, 7/100

This yields

0.4, 0.9, 0.07

8 0

3 years ago

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Shaded area

Andru [333]

Answer:

z-score value is 1.08

Step-by-step explanation:

The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Shaded area is 0.8599.

So, We are given the shaded area and we need to find the z-score

Looking at the Positive z-score value, which is added in the figure:

z = 1.08

So, z-score value is 1.08

6 0

3 years ago

Read 2 more answers

#18 I need help please!

vlabodo [156]

4+3=12 3*3*3==27 12-27=-15 5*{-15}=-75

3 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

4 years ago

Which choices are common factors of the numerator and denominator of the fraction below ? Check all that apply

melomori [17]

Answer:

1 and 7

Step-by-step explanation:

21: 1, 3, 7, 21

28: 1, 2, 4, 7, 14, 28

Common: 1, 7

4 0

4 years ago