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djverab [1.8K]
2 years ago
6

Four students wrote sequences during math class. Andre mc011-1.jpg Brenda mc011-2.jpg Camille mc011-3.jpg Doug mc011-4.jpg Which

student wrote a geometric sequence?
Mathematics
2 answers:
maks197457 [2]2 years ago
8 0
<span>A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

</span>The common ration is obtained by dividing the a term by the preceding term.

Given that f<span>our students wrote sequences during math class with
Andre writing -\frac{3}{4} ,\frac{3}{8} ,-\frac{3}{16} ,-\frac{3}{32} , . . .
Brenda writing </span>\frac{3}{4} ,-\frac{3}{8} ,\frac{3}{16} ,\frac{3}{32} , . . .
Camille writing \frac{3}{4} ,\frac{3}{8} ,-\frac{3}{16} ,-\frac{3}{32} , . . .
Doug writing \frac{3}{4} ,-\frac{3}{8} ,\frac{3}{16} ,-\frac{3}{32} , . . .

Notice that the common ratio for the four students is - \frac{1}{2}.

For Andre, the last term is wrong and hence his sequence is not a geometric sequence.
For Brenda, the last term is wrong and hence her sequence is not a geometric sequence.
For Camille, her sequence is not a geometric sequence.
For Doug, his sequence is a geometric sequence with a common ratio of - \frac{1}{2}.

Therefore, Doug wrote a geometric sequence.
mr_godi [17]2 years ago
3 0

Answer:

it was doug.

Step-by-step explanation:

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4. Suppose that Peculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply thr
Viktor [21]

Answer:

Peculiar purples would be more abundant

Step-by-step explanation:

Given that eculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply through a mechanism in which each single  bacterial cell splits into four. Time taken for one split is 12 m for I one and 10 minutes for 2nd

The function representing would be

i) P=P_0 (4)^{t/12} for I bacteria where t is no of minutes from start.

ii) P=P_0 (4)^{t/10} for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

a) Here P0 =3, time t = 60 minutes.

i) I bacteria P = 3(4)^{5} =3072

ii) II bacteria P = 3(4)^{4} =768

b) Since II is multiplying more we find that I type will be more abundant.

The difference in two hours would be

3(4)^{10}- 3(4)^{8} =2949120

c) i) P=P_0 (4)^{t/12} for I bacteria where t is no of minutes from start.

ii) P=P_0 (4)^{t/10} for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

d) At time 36 minutes we have t = 36

Peculiar purples would be

i) P=3 (4)^{36/12}=192

The rate may not be constant for a longer time.  Hence this may not be accurate.

e) when splits into 2, we get

P=P_o (2^t) where P0 is initial and t = interval of time

7 0
2 years ago
a previous analysis of paper boxes showed that the standard deviation of their lengths is 15 millimeters. A packers wishes to fi
vladimir1956 [14]

Answer:

864.36 boxes

Step-by-step explanation:

In the question above, we are given the following values,

Confidence interval 95%

Since we know the confidence interval, we can find the score.

Z score = 1.96

σ , Standards deviation = 15mm

Margin of error = 1 mm

The formula to use to solve the above question is given as:

No of boxes =[ (z score × standard deviation)/ margin of error]²

No of boxes = [(1.96 × 15)/1]²

= 864.36 boxes

Based on the options above, we can round it up to 97 boxes.

8 0
2 years ago
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an
Charra [1.4K]

1. Consider square ABCD. You know that  

A_{ABCD}=AD^2=200,

then

AB=BC=CD=AD=10\sqrt{2}.

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.

3. Consider trapezoid BFGC. Its area is

A_{BFGC}=\dfrac{FG+BC}{2}\cdot h, where h is the height of trapezoid and is equal to half of AB. Thus,

A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.

4.

A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.

5. Note that angles EGC and CGD are supplementary and

\sin \angle CGD=\sin \angle EGC.

Then

A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.

Answer: A_{CGD}=41.

8 0
2 years ago
A runner completes a 400-m race in 43.9 s. In a 100-m race, he finishes in 10.4 s. In which race was his speed faster?
Sveta_85 [38]

Answer:

100m race

Step-by-step explanation:

because 10.4 x 4 because 100m and 400m right 10.4x4=41.6 and 43.9 so obviously the 100m race is faster.

5 0
2 years ago
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if the sales tax rate is 7.25% in california, then how much would u pay in los angeles for a coat that cost $120.00
natima [27]

Answer:

The amount to be pay for coat is <u>$128.70</u>.

Step-by-step explanation:

Given:

The sales tax rate is 7.25% in California.

The cost for a coat in Los angeles is $120.00.

Now, to find amount to be paid.

Sales tax = 7.25%.

Cost = $120.00.

So, amount to be paid = $120 + 7.25% of $120.00.

=120+\frac{7.25}{100} \times 120.

=120+\frac{870}{100}.

=120+8.70

=\$128.70

Therefore, the amount to be pay for coat is $128.70.

7 0
2 years ago
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