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2 years ago

HELP QUICK PLEASE!!!!

Mathematics

2 answers:

kifflom [539]2 years ago

8 0

Answer:

It has 5 faces

Step-by-step explanation:

one on top (1), one on bottom (2), 2 on both sides (4), and one for the front so now you have 5 faces

just olya [345]2 years ago

3 0

Answer:

five

Step-by-step explanation:

when putting the image into a net you can count 5 sides

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Megan took out a loan for $500. The rate is 8%. how much does Megan owe in interest if she pays it back in 4.5 years

Arte-miy333 [17]

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3 years ago

Read 2 more answers

Some types of algae have the potential to cause damage to river ecosystems. Suppose the accompanying data on algae colony densit

Phantasy [73]

Answer:

$y=-2.95836 x +234.56159$

Step-by-step explanation:

We assume that th data is this one:

x: 50, 55, 50, 79, 44, 37, 70, 45, 49

y: 152, 48, 22, 35, 43, 171, 13, 185, 25

a) Compute the equation of the least-squares regression line. (Round your numerical values to five decimal places.)For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i =50+ 55+ 50+ 79+ 44+ 37+ 70+ 45+ 49=479$

$\sum_{i=1}^n y_i =152+ 48+ 22+ 35+ 43+ 171+ 13+ 185+ 25=694$

$\sum_{i=1}^n x^2_i =50^2 + 55^2 + 50^2 + 79^2 + 44^2 + 37^2 + 70^2 + 45^2 + 49^2=26897$

$\sum_{i=1}^n y^2_i =152^2 + 48^2 + 22^2 + 35^2 + 43^2 + 171^2 + 13^2 + 185^2 + 25^2=93226$

$\sum_{i=1}^n x_i y_i =50*152+ 55*48+ 50*22+ 79*35+ 44*43+ 37*171+ 70*13+ 45*185+ 49*25=32784$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=26897-\frac{479^2}{9}=1403.556$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=32784-\frac{479*694}{9}=-4152.22$

And the slope would be:

$m=-\frac{-4152.222}{1403.556}=-2.95836$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{479}{9}=53.222$

$\bar y= \frac{\sum y_i}{n}=\frac{694}{9}=77.111$

And we can find the intercept using this:

$b=\bar y -m \bar x=77.1111111-(-2.95836*53.22222222)=234.56159$

So the line would be given by:

$y=-2.95836 x +234.56159$

7 0

2 years ago

Mealworms are a healthy food for wild song birds. Adam buys a 3 oz container of mealworms for $7.35. Marco buys a 5 oz container

Artemon [7]

Answer: 5 oz

Step-by-step explanation:

you get more and for the price of 7.35 it’s not worth only getting three oz

3 0

2 years ago

Which is the value of this expression when a=5 and k=-2

Zinaida [17]

Answer:

Option C is correct.

Step-by-step explanation:

We are given the expression:

$(\frac{3^2a^{-2}}{3a^{-1}})^k$

The value of a =5 and k = -2

Putting the values and solving

$=(\frac{3^2*5^{-2}}{3*5^{-1}})^-2\\=(\frac{3^{2-1}}{5^{-1+2}})^-2\\=(\frac{3^{1}}{5^{1}})^-2\\\\=(\frac{3}{5})^-2\\if \,\,a^{-1} \,\,then\,\, 1/a\\=\frac{(3)^{-2}}{(5)^{-2}}\\ Can\,\,be\,\,written\,\,as\\\\=\frac{(5)^{2}}{(3)^{2}} \\=\frac{25}{9}$

Option C is correct.

3 0

3 years ago

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

$Z_{1-\alpha /2}= Z_{0.95}= 1.648$

0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

0.48±1.648* $\sqrt{\frac{0.48*0.52}{1000} }$

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: $\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015$

Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0

3 years ago