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3 years ago

Ind the 7th term of the geometric sequence with the given terms.

1 answer:

3 0

To find the seventh term of the geometric sequence, divide first the fifth term by the fourth term to obtain the common ratio. -40 divide by -8 is 5. Then, multiply a5 by the common ratio, 5 twice or a4 by 5 three times. Both solutions will give, an a7 equal to -1000. Thus, a7 is -1000.

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Find the area of the trapezoid. Answer without units.

Leto [7]

Answer:

180

Step-by-step explanation:

Since the type of trapezoid is not listed, I cannot really explain, sorry :/

5 0

3 years ago

I need help please i would appreciate it

Reptile [31]

Answer: 5/3

i hope that’s correct :)

3 0

3 years ago

Read 2 more answers

Two buses leave towns 304 miles apart at the same time and travel toward each other. One bus travels 14 mih slower than the othe

8090 [49]

Answer:

The faster bus moves at 83mi/h and the slower one moves at 69mi/h.

Step-by-step explanation:

Let's define:

R₁ = rate of bus 1, this is the faster one.

R₂ = rate of bus 2, this is the slower one.

We know that one bus travels 14mi/h slower, then:

R₂ = R₁ - 14mi/h.

Now we know that:

Distance = Speed*Time.

If we add the distances that both busses travel in 2 hours, it should be equal to the initial distance between the buses, then:

R₁*2h + R₂*2h = 304 mi

Then we have the two equations:

R₂ = R₁ - 14mi/h

R₁*2h + R₂*2h = 304 mi

The first step is to replace the first equation in the second one, to get:

R₁*2h + (R₁ - 14mi/h)*2h = 304 mi

And now we can solve this for R₁.

R₁*2h + R₁*2h - 14mi/h*2h = 304 mi

R₁*4h - 28mi = 304mi

R₁*4h = 304mi + 28mi = 332mi

R₁ = 332mi/4h = 83mi/h

The faster bus moves at 83mi/h

And we know that the slower one moves at 14mi/h slower than this, then:

R₂ = R₁ - 14mi/h = 83mi/h - 14mi/h = 69 mi/h

7 0

2 years ago

The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric

Papessa [141]

Let $X$ denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately $z=-0.2533$ , which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately $z=2.0538$ , so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .

7 0

3 years ago

Out of the thirty

labwork [276]

Answer:

She answered 76 percent of the question correct.

Step-by-step explanation:

Convert fraction (ratio) 23 / 30 Answer: 76.6%

8 0

3 years ago