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Mashutka [201]
3 years ago
14

What is the image of (-2,9)after a dilation by a scale factor of 5 centered at the origin?

Mathematics
1 answer:
asambeis [7]3 years ago
7 0

Answer:

(- 10, 45 )

Step-by-step explanation:

Since the dilatation is centred at the origin, then multiply each of the coordinates by the scale factor, that is

(- 2, 9 ) → (- 2 × 5, 9 × 5 ) → (- 10, 45 )

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Compare and contrast expressions equations and inequalities
babymother [125]
Equations can have one solution, but inequalities have infinite solutions.

Hope this helps!
4 0
3 years ago
A distribution of values is normal with a mean of 220 and a standard deviation of 13. From this distribution, you are drawing sa
Paraphin [41]

Answer:

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributied random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 220, \sigma = 13, n = 35, s = \frac{13}{\sqrt{35}} = 2.1974

Find the interval containing the middle-most 48% of sample means:

50 - 48/2 = 26th percentile to 50 + 48/2 = 74th percentile. So

74th percentile

value of X when Z has a pvalue of 0.74. So X when Z = 0.643.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

0.643 = \frac{X - 220}{2.1974}

X - 220 = 0.643*2.1974

X = 221.41

26th percentile

Value of X when Z has a pvalue of 0.26. So X when Z = -0.643

Z = \frac{X - \mu}{s}

-0.643 = \frac{X - 220}{2.1974}

X - 220 = -0.643*2.1974

X = 218.59

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

5 0
3 years ago
Suppose that the cost of producing x tablets is defined by c(x)= 200 + 10x + 0.2x squared* where x represents the number if tabl
sammy [17]

we are given

C(x)=200+10x+0.2x^2

part-A:

Since, x is number of tablets

C(x) is cost of producing x tablets

so, vertical box is C(x)

so, we write in vertical box is "cost of producing x tablets"

Horizontal box is x

so, we write in horizontal box is "number of tablets"

part-B:

we have to find average on [a,b]

we can use formula

\frac{C(b)-C(a)}{b-a}

we are given point as

a: (15 , 395)

a=15 and C(15)=395

b: (20, 480)

b=20 , C(20)=480

now, we can plug values

=\frac{480-395}{20-15}

=17...........Answer

part-c:

we have to find average on [b,c]

we can use formula

\frac{C(c)-C(b)}{c-b}

we are given point as

b: (20, 480)

b=20 , C(20)=480

c:(25,575)

c=25 , C(c)=575

now, we can plug values

=\frac{575-480}{25-20}

=19..............Answer


5 0
3 years ago
HELP WHATS THE CONSTANT OF PROPORTIONALITY<br> sodas: 2 3 4 5 <br> cost: $2.50 $3.75 $5.00 $6.25
musickatia [10]

Answer:

its 1.25

Step-by-step explanation:

3 0
2 years ago
Find the number that we can place in the box, so that the resulting quadratic is the square of a binomial.
Arturiano [62]

We need a number to complete

(x+c)^2=x^2+2xc+c^2

Since we have

2xc=\dfrac{4}{5}x \iff 2c=\dfrac{4}{5}\iff c=\dfrac{4}{10}=\dfrac{2}{5}

And so the expanded square is

\left( x+\dfrac{2}{5}\right)^2=x^2+\dfrac{4}{5}x+\dfrac{4}{25}

6 0
3 years ago
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