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Morgarella [4.7K]
3 years ago
13

Compare the y-intercepts and the rates of change of the following items.

Mathematics
1 answer:
marysya [2.9K]3 years ago
8 0

Answer:

The answer is C: They are all the same.

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In a sale, normal prices are reduced by 17%. The normal price of a washing machine is reduced by £42.50
Misha Larkins [42]

Answer: The original price of washing machine is $250.

Step-by-step explanation:

Since we have given that

Rate of decline in normal prices = 17%

Amount of normal price reduced by = $42.50

Let the original price be x

According to question,

\frac{17}{100}\times x=42.50\\\\x=\frac{100}{17}\times 42.50\\\\x=\$250

Hence, The original price of washing machine is $250.

8 0
3 years ago
HELP I’m almost out of time ( it’s not Y=2x+4)
STALIN [3.7K]

Answer:

y=2/5

Step-by-step explanation:

_+;3_;3(_+3++_83 I got you man

8 0
3 years ago
If m<13=m<15=m<7, what conclusions can you make about lines a, b, c, and d?
Strike441 [17]

Answer:

option D

Lines a and b are parallel and lines c and d are parallel.

Step-by-step explanation:

Given in the question four lines a , b, c, d.

<h3>Prove one</h3>

If two parallel lines (a,b) are cut by a transversal(d), then corresponding angles m<7 and m<15 are congruent.

They are know as corresponding angles.

Hence lines a and b are parallel.

<h3>Prove two</h3>

If two parallel lines (c,d) are cut by a transversal(d), then corresponding angles m<13 and m<15 are congruent.

They are know as corresponding angles

Hence lines c and d are parallel

5 0
3 years ago
I need help with this math qustoin 2(2+455×330)÷3×2 and anyone bored
____ [38]
Solve what is in the parenthesis first. Then multiply that by 2 then divide that by 3 then multiply that by 2. know your order of operations. 
6 0
3 years ago
Read 2 more answers
The principal randomly selected six students to take an aptitude test. Their scores were: 81.6 72.0 81.1 86.4 70.2 83.1 Determin
tatyana61 [14]

Answer:

84.38, 73.74

Step-by-step explanation:

score given 81.6, 72.0, 81.1, 86.4, 70.2, 83.1

sample size (n) = 6

mean = \dfrac{81.6+ 72.0+ 81.1+ 86.4+ 70.2+ 83.1}{6}

mean = 79.06

standard deviation

\sigma =\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}

\sigma =\sqrt{\frac{ (81.6-79.06)^2+(72-79.06)^2+(86.4-79.06)^2+(70.2-79.06)^2+(81.1-79.06)^2+(83.1-79.06)^2}{6-1}}

σ = 6.47

level of significance (α) = 1 - 90% = 10%

confidence interval

\bar{x} \pm t_{\alpha}(\frac{S}{\sqrt{n}})\\79.06 \pm 2.015(\frac{6.47}{\sqrt{6} })

=79.06 ± 5.32

= 84.38, 73.74

8 0
3 years ago
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