PLEASEEEE HELP ME

Can someone helpp me with this i will give brainlest :)

marta [7]

Answer:

Step-by-step explanation:

1slice=280 and 6 slices so 280x6 so 0x6=0 then 80x6=6+6x4=48x10=480
then 200x6= 6x2=12x100=1200

add all 0+480+1200=1680

sry i'm not used to showing work i usually do it all in my head

8 0

1 year ago

The lengths of the four sides of a quadrilateral (in centimeters) are consecutive

My name is Ann [436]

The longest of the four side lengths is 24 cm

<h3>How to find the value of the longest of the four side lengths?</h3>

The given parameters are:

Perimeter, P = 90 cm

The lengths are consecutive integers

Let the smallest be

Smallest = x

So, we have

P = x + x + 1 + x + 2 + x + 3

Evaluate the like terms

P = 4x + 6

Substitute P = 90 cm in P = 4x + 6

4x + 6 = 90

Subtract 6 from both sides

4x = 84

Divide by 4

x = 21

The longest side is

Longest = x + 3

This gives

Longest = 21 + 3

Evaluate

Longest = 24

Hence, the longest of the four side lengths is 24 cm

Read more about perimeter at:

brainly.com/question/19819849

#SPJ1

6 0

2 years ago

What is the solution of -3 = x + 5

zhenek [66]

-8 because -8 +5=-3 that is the equation used to get the solution

7 0

3 years ago

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that

kompoz [17]

Answer:

a)

The probability that a single score drawn at random will be greater than 110

P( X > 110) = 0.1587

b)

The probability that a sample of 25 scores will have a mean greater than 105

P( x> 105) = 0.0062

c)

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

Step-by-step explanation:

a)

Given mean of the Normal distribution 'μ' = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

The probability that a single score drawn at random will be greater than 110

P( X > 110) = P( Z >1)

= 1 - P( Z < 1)

= 1 - ( 0.5 +A(1))

= 0.5 - A(1)

= 0.5 -0.3413

= 0.1587

b)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

The probability that a single score drawn at random will be greater than 110

P( x> 105) = P( z > 2.5)

= 1 - P( Z< 2.5)

= 1 - ( 0.5 + A( 2.5))

= 0.5 - A ( 2.5)

= 0.5 - 0.4938

= 0.0062

The probability that a single score drawn at random will be greater than 105

P( x> 105) = 0.0062

c)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

The probability that a single score drawn at random will have a mean greater than 105

P( x> 105) = P( z > 4)

= 1 - P( Z< 4)

= 1 - ( 0.5 + A( 4))

= 0.5 - A ( 4)

= 0.5 - 0.498

= 0.002

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

Let x₁ = 95

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

Let x₂ = 105

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

= P(z≤2) - P(z≤-2)

= 0.5 + A( 2) - ( 0.5 - A( -2))

= A( 2) + A(-2) (∵A(-2) =A(2)

= A( 2) + A(2)

= 2 × A(2)

= 2×0.4772

= 0.9544

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

7 0

3 years ago

Giving out brainlst to whoever can answer this

Harrizon [31]

100 - 5d + 10d = 130

5 x 20 = 100
5 x -d = -5d

3 0

4 years ago