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levacccp [35]
3 years ago
13

Teresa graphs the following 3 equations: y=2^x, y=x^2+2, and y=2x^2. She says that the graph of y=2^x will eventually surpass bo

th of the other graphs. Is Teresa correct? Why or why not?
A.Teresa is correct.
The graph of y=2xy=2x grows at an increasingly increasing rate, but the graphs of y=x2+2y=x2+2 and y=2x2y=2x2 both grow at a constantly increasing rate.
Therefore, the graph of y=2xy=2x will eventually surpass both of the other graphs.

B. Teresa is not correct.
The graph of y=2xy=2x grows at an increasing rate and will eventually surpass the graph of y=x2+2y=x2+2.
However, it will never surpass the graph of y=2x2y=2x2 because the yy-value is always twice the value of x2x2.

C. Teresa is not correct.
The graph of y=2x2y=2x2 already intersected and surpassed the graph of y=2xy=2x at x=1x=1.
Once a graph has surpassed another graph, the other graph will never be higher.

Mathematics
2 answers:
makvit [3.9K]3 years ago
6 0
A) Teresa is correct; y=2ˣ grows at an increasingly increasing rate, while the other two grow at a constantly increasing rate. This means y=2ˣ will surpass the other two.
ryzh [129]3 years ago
5 0

Answer:

The graph of y=2^{x} will eventually surpass both of the other graphs.


Step-by-step explanation:

Given three equations

y=2^{x} , y=x^{2} +2, y=2x^{2}

Now, Teresa says that the graph of y=2^{x} will eventually surpass both of the other graphs. we have to find is she correct whether the graph surpass which means there exist any value of x such that y=2^{x} greater than both other value of y.If we can not find any value of x  then graph does not surpass the other.As we know the exponential function grows increasingly increasing rate than that of polynomial function. In the graph shown it is clear that the graph of  [tex]y=2^{x}  grows at an increasingly increasing rate, but the graphs of  y=x^{2} +2, y=2x^{2} both grow at a constantly increasing rate.


Therefore, the graph of y=2^{x} will eventually surpass both of the other graphs.




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3 years ago
The vector v⃗ in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis.
allsm [11]

The vector v in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis is: (3/√2, 3√2) and The vector w in 3-space of length 1 lying in the yz-plane pointing upward at an angle of 2π/3 measured from the positive y-axis is: (0, -1/2, √3/2).

<h3>Vector</h3>

a. Vector (v)

Vector (v)=v (cos Ф, sin Ф)

V=1 while the counterclockwise angle that is measured from positive x=Ф=π/4

Hence:

Vector=3(cos π/4, sinπ/4)

Vector=(3/√2, 3√2)

b. Vector w:

Vector w=1(0, cos 2π/3, sin2π/3)

Vector w=(0, -1/2, √3/2)

Therefore the vector v in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis is: (3/√2, 3√2) and The vector ws: (0, -1/2, √3/2).

The complete question is:

Resolve the following vectors into components:

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5 0
2 years ago
A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain
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Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55

The mean of this distribution is:

\mu=np=10*0.6=6

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8

The probability of having between 4 and 8 customers choosing the oversize version is:

P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328

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3 years ago
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Answer:

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Step-by-step explanation:

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