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Brums [2.3K]
2 years ago
12

If jl= 10x-2 JK = 5x-8 and KL = 7x-12 find KL

Mathematics
1 answer:
natta225 [31]2 years ago
8 0
JK+KL=JL
(5x-8)+7x-12=10x-2
(add common terms together)
12x-20=10x-2
(get the term with x alone on one side of the equation)
2x=18
(divide by 2 to get x alone)
x=9

Find KL:
KL=7(9)-12
KL=51
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The selling price of a box of crackers is $1.75 You mark the crackers up to $2.54 . What is the markup percentage?
Aleks04 [339]

The markup percentage is 45.14%

Step-by-step explanation:

The given is:

  • The selling price of a box of crackers is $1.75
  • You mark the crackers up to $2.54

We need to find the markup percentage

The markup percentage = \frac{New-Old}{old} × 100%

∵ The selling price of a box of crackers is $1.75

∴ Old = 1.75

∵ You mark the crackers up to $2.54

∴ New = 2.54

- Substitute these values in the rule above

∵ The markup percentage = \frac{2.54-1.75}{1.75} × 100%

∴ The markup percentage = \frac{0.79}{1.75} × 100%

∴ The markup percentage = 0.4514 × 100%

∴ The markup percentage = 45.14%

The markup percentage is 45.14%

Learn more:

You can learn more about percentage in brainly.com/question/1834017

#LearnwithBrainly

8 0
2 years ago
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Colin gets pic 'n mix when at the cinema and makes his bag up with 3 types of sweet. He pick:
Kisachek [45]

Answer: 2/3

Explanation: Assume that the number of smarties he picked is c.

Now, we are given that:

1- He picked 3 times as many cola bottles as smarties. This means that:

Number of cola bottles he picked = 3s

2- He picked twice as many smarties as marshmallows. This means that:

number of marshmallows he picked = 0.5s

Now to get the proportion o f cola bottles, we will divide the number of cola bottles by the total number of sweets as follows:

proportion of cola bottles =  

Hope this helps :)

4 0
1 year ago
ΔABC is dilated using a scale factor of 12 to produce ΔA'B'C'. Select all of the statements that apply to the transformation.
GalinKa [24]

Answer:

Options (3) and (6)

Step-by-step explanation:

ΔABC is a dilated using a scale factor of \frac{1}{2} to produce image triangle ΔA'B'C'.

Since, dilation is a rigid transformation,

Angles of both the triangles will be unchanged or congruent.

m∠A = m∠A' and m∠B = m∠B'

Since, sides of ΔA'B'C' = \frac{1}{2} of the sides of ΔABC

Area of ΔA'B'C' = \frac{1}{2}(\text{Area of triangle ABC})

Area of ΔABC > Area of ΔA'B'C'

Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.

ΔABC ~ ΔA'B'C'

Therefore, Option (3) and Option (6) are the correct options.

3 0
1 year ago
Samara is adjusting a satellite because she finds it is not focusing the income radio waves perfectly. The shape of her satellit
gogolik [260]

Answer:

\boxed{\text{(6, 3)}}

Step-by-step explanation:

The conic form of the equation for a sideways parabola is

(y - k)² = 4p(x - h)

The focus is at (h + p, k)

The equation of Samara's parabola is

(y - 3)² = 8(x - 4)

h = 4

p = 8/4 = 2

k = 3

h + p = 6

So, the focus point of the satellite dish is at

\boxed{\textbf{(6, 3)}}

8 0
1 year ago
Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class p
Dima020 [189]

Answer:

a) p-hat (sampling distribution of sample proportions)

b) Symmetric

c) σ=0.058

d) Standard error

e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

Step-by-step explanation:

a) This distribution is called the <em>sampling distribution of sample proportions</em> <em>(p-hat)</em>.

b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.

This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.

c) The variability of this distribution, represented by the standard error, is:

\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058

d) The formal name is Standard error.

e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

0 0
2 years ago
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