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Ivanshal [37]
3 years ago
14

Whats the answer to this problem/how to figure it out?!

Mathematics
2 answers:
Evgen [1.6K]3 years ago
7 0

Answer:

<h2>330</h2>

Step-by-step explanation:

6x^2 +4x +8\\x =7\\\\6(7)^2 +4(7) +8\\\mathrm{Follow\:the\:PEMDAS\:order\:of\:operations}\\\\\mathrm{Calculate\:exponents}\:\left(7\right)^2\::\quad 49\\\\=6\cdot \:49+4\left(7\right)+8\\\\\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:6\cdot \:49\::\quad 294\\\\=294+4\left(7\right)+8\\\\\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:4\left(7\right)\::\quad 28\\\\=294+28+8\\\\\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:294+28+8\:\\:\quad 330

Readme [11.4K]3 years ago
5 0

6x^2+4x+8

6x7^2+4x+8

blah  blah blah

ANSWER

330

pls say if im right

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This year the CDC reported that 30% of adults received their flu shot. Of those adults who received their flu shot,
Vlad [161]

Using conditional probability, it is found that there is a 0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Person has the flu.
  • Event B: Person got the flu shot.

The percentages associated with getting the flu are:

  • 20% of 30%(got the shot).
  • 65% of 70%(did not get the shot).

Hence:

P(A) = 0.2(0.3) + 0.65(0.7) = 0.515

The probability of both having the flu and getting the shot is:

P(A \cap B) = 0.2(0.3) = 0.06

Hence, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.515} = 0.1165

0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

To learn more about conditional probability, you can take a look at brainly.com/question/14398287

7 0
2 years ago
Find the slope of the line that passes through (1, 8) and (10, 1).
leonid [27]

Answer:

m =  \frac{1 - 8}{10 - 1}  =   - \frac{  7}{9}  \\

Slope(gradient)=

-\frac{7}{9}

Good luck!

Intelligent Muslim,

From Uzbekistan.

8 0
3 years ago
Read 2 more answers
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!

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4 years ago
Fraction that is less than 5/6 and has a denominator of 8
Evgesh-ka [11]
See 5/6 is obviously more than 1/2 so 4/8 works
also 3/8 or 2/8 or 1/8

if you want all fractions then
5/6=x/8
times 24 both sides
20=3x
divide 3
6 and 2/3=x
basically x/8 such that x<6 and 2/3


examples
6/8
5/8
4/8
3/8
2/8
1/8
I would use the odd ones since the even ones simplify to non-eight denomenators
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3 years ago
A dog eats 2/5
serious [3.7K]
The answer for your dog would be 2-4 7.8 that your answer your welcome
4 0
3 years ago
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