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LuckyWell [14K]
3 years ago
11

Kiran's mother gets a restaurant bill for $25. She has a coupon for 15% off. After the

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

$25.50

Step-by-step explanation:

So we have a 15% off coupon. We can multiply 25×.85 to find our new price.

Our total is now $21.25

Now to find the tip do $21.25 +(.2 × 21.25)

*The .2 is 20 percent of our discount price.*

We then get $25.50

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Consider the parabola given by the equation: f(x) = 2x2 + 14.0 – 4
Softa [21]

Answer:

Step-by-step explanation:

f(x)  = 2x² + 14x - 4

= 2(x² + 7x) - 4

= 2(x² +7x + 3.5²) - 2(3.5²) - 4

= 2(x+3.5)² - 28.5

vertex (3.5, -28.5) = (7/2, -57/2)

The vertical intercept is the y-intercept, i.e., f(0) = -4.

The x-intercepts are the values of x for which y=0.

2x² + 14x - 4  = 0

x = [-14±√(14²-4(2)(-4))]/[2(2)] = [-7±√57]/2 ≅ -7.27, 0.27

5 0
2 years ago
The ages of armadillos are normally distributed, with a mean of 14 years and a standard deviation of 1.2. Approximately what per
vovangra [49]

Answer:

Percentage of armadillos between 13 and 17 years = 79.052%f using Standard Normal Distribution Tables

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu = 14, SD = 1.2 we have z(x) by using z(x) = (x - Mu)/SD as under:

Approach 1 using Standard Normal Distribution Table:

z for x=17: z(17) = (17-14)/1.2 gives us z(17) = 2.5

z for x=13: z(13) = (13-14)/1.2 gives us z(13) = -0.83

Afterwards using Normal Distribution Tables we find the probabilities as under:

P(17) using z(17) = 2.5 gives us P(17) = 99.379%

Similarly we have:

P(13) using z(13) = -0.83 gives us P(13) = 20.327%

Finally in order to find out the probability between 17 & 13 years we have:

Percentage of armadillos between 13 and 17 years = P(17) - P(13) = 99.379% - 20.327% = 79.052%

The standard normal distribution table is being attached for yours easiness.

Approach 2 using Excel or Google Sheets:

P(17) = norm.dist(17,14,1.2,1)

P(13) = norm.dist(13,14,1.2,1)

Percentage of armadillos between 13 and 17 years = { P(17) - P(13) } * 100

Download pdf
4 0
3 years ago
Read 2 more answers
For which pair of function is (g*f)(a)=|a|-2
tia_tia [17]
None of the pairs will deliver (g×f)(a). If you intend (g∘f)(a), then ...

... selection 3 is appropriate.

5 0
3 years ago
Read 2 more answers
Help helpp pleaseee
Len [333]

Answer:

3 is the right I think idonk

8 0
3 years ago
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311
Bingel [31]

Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

Z_{1-\alpha /2}= Z_{0.975}= 1.965

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

5 0
3 years ago
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