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3 years ago

Find the sum of the finite geometric series. 4 20 100 ... 312,500 the sum of the geometric series is .

1 answer:

5 0

What is the 9th term of the geometric sequence 4, -20, 100, ...
First term, a = 4 
common ratio, r = –20 / 4 = –5 
..... ..... ..... ..... = 100 / –20 = –5 
9th term = 4 ( –5 ) ^ ( 9 - 1 ) 
..... ...... = 4 ( –5 ) ^ 8 
..... ...... = 4 ( 390,625 ) 
..... ...... = 1,562,500

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Hello, I need help with a math question, please tell me if I am right. *Refer to the photo*

Trava [24]

As I can tell the tables are right but it would be a function since none of the x's repeat

5 0

2 years ago

the perimeter of the floor of a room is 18 metre and its height is 3 cm what is the area of 4 walls of the room

Lina20 [59]

Note: The height of the room must be 3 m instead of 3 cm because 3 cm is too small and it cannot be the height of a room.

Given:

Perimeter of the floor of a room = 18 metre

Height of the room = 3 metre

To find:

The area of 4 walls of the room.

Solution:

We know that, the area of 4 walls of the room is the curved surface area of the cuboid room.

The curved surface area of the cuboid is

$C.S.A.=2h(l+b)$

Where, h is height, l is length and b is breadth.

Perimeter of the rectangular base is 2(l+b). So,

$C.S.A.=\text{Perimeter of the base} \times h$

Putting the given values, we get

$C.S.A.=18\times 3$

$C.S.A.=54$

Therefore, the area of 4 walls of the room is 54 sq. metres.

7 0

3 years ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$

7 0

3 years ago

Help thanks 7.04 love u guys

Ugo [173]

Answer:

D. $-\frac{3}{4}$

Step-by-step explanation:

The given equation is

$4y=8x-3$

When we divide through by 4 we get;

$y=2x-\frac{3}{4}$

Comparing to

y=mx+c,

The y-intercept is $-\frac{3}{4}$

The corrrect choice is D.

3 0

3 years ago

Read 2 more answers

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = $\frac{354}{1219}$

n = sample of US adults = 1291

p = population proportion of adults

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

significance level are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

= [ $\frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ , $\frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

5 0

3 years ago