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3 years ago

Write the first 4 terms for this sequence t(n)=4.5n-8

2 answers:

6 0

Evaluate t(1),t(2),t(3),t(4)

t(1)=4.5(1)-8=4.5-8=-3.6
t(2)=4.5(2)-8=9-8=1
t(3)=4.5(3)-8=13.5-8=5.5
t(4)=4.5(4)-8=18-8=10

the first 4 terms are
-3.6,1,5.5, 10

Anon25 [30]3 years ago

3 0

Simplifying t(n) = 4.5n + -8 Multiply t * n nt = 4.5n + -8 Reorder the terms: nt = -8 + 4.5n Solving nt = -8 + 4.5n Solving for variable 'n'. Move all terms containing n to the left, all other terms to the right. Add '-4.5n' to each side of the equation. -4.5n + nt = -8 + 4.5n + -4.5n Combine like terms: 4.5n + -4.5n = 0.0 -4.5n + nt = -8 + 0.0 -4.5n + nt = -8 <span>Reorder the terms: 8 + -4.5n + nt = -8 + 8 Combine like terms: -8 + 8 = 0 8 + -4.5n + nt = 0 The solution to this equation could not be determined.</span>

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A class survey in a large class for first-year college students asked, “About how many minutes do you study on a typical weeknig

Anika [276]

Answer: (110.22, 125.78)

Step-by-step explanation:

The confidence interval for the population mean is given by :-

$\mu\ \pm z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size = 463

$\mu=118\text{ minutes}$

$\sigma=65\text{ minutes}$

Significance level : $\alpha=1-0.99=0.01$

Critical value : $z_{\alpha/2}=z_{0.005}=\pm2.576$

We assume that the population is normally distributed.

Now, the 90% confidence interval for the population mean will be :-

$118\ \pm\ 2.576\times\dfrac{65}{\sqrt{463}} \\\\\approx118\pm7.78=(118-7.78\ ,\ 118+7.78)=(110.22,\ 125.78)$

Hence, 99% confidence interval for the mean study time of all first-year students = (110.22, 125.78)

3 0

3 years ago

The density of a certain material is such that it weighs 9 kilograms for every 7 cubic feet of volume. Express this density in p

Vadim26 [7]

Answer:

1.641× 10^-3 pounds/ cubic inch

Step-by-step explanation:

Density= mass / volume

Mass of the material= 9 kilograms

Volume= 7 cubic feet

Density= 9/7= 1.286 kg/ ft^3

Density= 1.286 kg/ ft^3

To Convert to pounds per inch

1 kg/ ft^3 = 0.00128 pound / cubic inch

1.286 kg/ ft^3 = X pounds/ cubic inch

X=( 1.286 × 0.00128)

= 0.0016407087 pounds/ cubic inch

= 1.641× 10^-3 pounds/ cubic inch

Hence,this density in pounds per inch is 1.641× 10^-3 pounds/ cubic inch

5 0

2 years ago

Application

user100 [1]

He needs to run with approximately 6429 for a distance of 9 km

<h3>How to determine the number of strides?</h3>

The given parameters are:

Length of stride = 1.4 m

Distance for marathon = 9 km

The number of strides needed is then calculated as:

Number of stride = Distance for marathon/Length of stride

Substitute the known values in the above equation

Number of stride = 9km/1.4m

Convert km to m

Number of stride = 9000m/1.4m

Evaluate the quotient

Number of stride = 6428.57143

Approximate the estimate

Number of stride = 6429

Hence, he needs to run with approximately 6429 for a distance of 9 km

Read more about quotients at:

brainly.com/question/8952483

#SPJ1

6 0

2 years ago

Consider the given function and the given interval. f$x$ = (x - 7)**2 text(, ) [5 text(, ) 11] (a) Find the average value fave

Jet001 [13]

a. $f$ has an average value on [5, 11] of

$f_{\rm ave}=\displaystyle\frac1{11-5}\int_5^{11}(x-7)^2\,\mathrm dx=\frac{(x-7)^3}{18}\bigg|_5^{11}=\frac{4^3-(-2)^3}{18}=4$

b. The mean value theorem guarantees the existence of $c\in(5,11)$ such that $f(c)=f_{\rm ave}$ . This happens for

$(c-7)^2=4\implies c-7=\pm2\implies c=9\text{ or }c=5$

8 0

3 years ago

Can somebody help me with this math question?

kap26 [50]

Answer:Hk and IJ

Step-by-step explanation:

They are parallel to eachother, hope this helped!

<!> Brainliest is appreciated! <!>

6 0

3 years ago