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geniusboy [140]
2 years ago
13

Maddie is reading a graph of equivalent ratios. One point on the graph is (4, 5). Maddie says (10, 8) is also on the graph. Is M

addie correct? Explain why or why not.
Mathematics
2 answers:
dmitriy555 [2]2 years ago
6 0
Yes and no, (4, 5) multiplied by 2 is (10, 8) but reversed. It depends on how you look at it because (10, 8) would be on the other side of the graph than (8, 10) which is the actually equivalent.
Vesna [10]2 years ago
5 0

Answer: Yes and no, (4, 5) multiplied by 2 is (10, 8) but reversed. It depends on how you look at it because (10, 8) would be on the other side of the graph than (8, 10) which is the actually equivalent.

Step-by-step explanation: hope it helped

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How many ounces of pure nickel must be added to 150 ounces of alloy 70% pure to make an alloy which is 85% pure?
galben [10]

Answer:

150 oz.

Step-by-step explanation:

There are already 150 ounces of alloy of nickel.

Of this 150 oz, 70% is pure i.e. nickel content = 150(0.7) = 105 oz

Now available is

Nickel       Other metals

105                  45

Let x oz of pure nickel is added.

Then new alloy will have 105+x oz nickel in total of 150+x oz.

Percentage pure = \frac{105+x}{150+x} =85%

Simplify to get

\frac{105+x}{150+x} =\frac{85}{100} \\\\Cross multiply:\\85(150+x) =100(105+x)\\12750+85x =10500+100x\\100x-85x = 12750-10500\\15x = 2250\\x = 150

Hence answer is 150 oz should be added.

6 0
2 years ago
I need to solve this equation: x= r-h/ y<br> for h and then for r
ololo11 [35]
X= r-h/y
h= xy-r/-1
r= xy+h
7 0
2 years ago
Read 2 more answers
A fabric store sells two types of ribbon. One customer buys 3 rolls of the lace ribbon and 2 rolls of the satin ribbon and has a
stepan [7]
<span>Let L be the number of yards on a roll of lace ribbon. Let S be the number of yards on a roll of satin ribbon. We can set up two equations. equation 1: 3L + 2S = 120 yards equation 2: 2L + 4S = 160 yards We can multiply (equation 1) by 2 and subtract (equation 2). equation 1: 6L + 4S = 240 yards equation 2: 2L + 4S = 160 yards 4L = 80 yards L = 20 yards equation 1: 3L + 2S = 120 yards 3(20 yards) + 2S = 120 yards 2S = 60 yards S = 30 yards There are 20 yards on a roll of lace ribbon. There are 30 yards on a roll of satin ribbon.</span>
7 0
2 years ago
Minato drove 390 miles. Part of the drive was along local roads, where his average speed was 20 mph, and the rest was along a hi
pashok25 [27]

Answer:

45 miles.

Step-by-step explanation:

Given that the:

Total distance covered = 390 miles

Total time = 8 hours

Let the distance covered along the local way = L

And the distance covered along the highway = H

Along with local way,

Speed = distance/ time

20 = L / T

T = L /20 .... (1)

Along the highway,

Distance covered H = 390 - L

Let the time = t

Speed = distance/time

60 = (390 - L)/t

t = ( 390 - L)/60

But total time = T + t

That is

8 = L/20 + (390 - L)/60

The LCM at right hand side will be 60

8 = ( 3L + 390 - L )/60

Cross multiply

480 = 2L + 390

Collect the like terms

2L = 480 - 390

2L = 90

L = 90/2

L = 45 miles.

Therefore, the distance Minato drive along local roads is 45 miles

4 0
2 years ago
"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt
Westkost [7]

Answer:

a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

d) P(\bar X >28.25)=0.9997=99.97\%

e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

3 0
2 years ago
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