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lorasvet [3.4K]
2 years ago
8

PLEASE HELP!! 10 POINTS!! WILL MARK BRAINLIEST!!

Mathematics
1 answer:
Hoochie [10]2 years ago
7 0

An octagon has 8 sides. Each side is 2a + 1 units.

8(2a + 1)

= 16a + 8

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5. Write the equation of the line graphed below.
Pepsi [2]

Answer:

y=2x+3

Step-by-step explanation:the rise over run is 2/1 indicating that the slope is 2. the line also passes through 3 which means the y intercept is 3.  

7 0
2 years ago
What is the value of x in this equation: 4x + 5 = 3x + 4? Solve the equation using the algebra tiles.
Snowcat [4.5K]

Answer:

x = -1

Step-by-step explanation:

4x + 5 = 3x + 4

      -5           -5

4x =    3x -1

-3x =   -3x

1x = -1

x = -1

4 0
2 years ago
Read 2 more answers
A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0
2 years ago
Coupons driving visits. A store randomly samples 603 shoppers over the course of a year and nds that 142 of them made their visi
s2008m [1.1K]

Answer:

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 603, \pi = \frac{142}{603} = 0.2355

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

7 0
2 years ago
If 6j-5k =11 and 5j-6k =22, then what is the value of 2j+2k??
dsp73

Answer:

-22

Step-by-step explanation:

6j-5k=11

5j-6k=22

-----------------

5(6j-5k)=5(11)

-6(5j-6k)=-6(22)

-----------------------

30j-25k=55

-30j+36k=-132

-----------------------

11k=-77

k=-77/11

k=-7

6j-5(-7)=11

6j+35=11

6j=11-35

6j=-24

j=-24/6

j=-4

------------------

2j+2k=2(-4)+2(-7)=-8-14=-22

8 0
2 years ago
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