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3 years ago

Find the solution set of this inequality. select the correct graph. |-4x+14|>6

1 answer:

5 0

Split the inequality into two:
(-4x+14)>6 & -(-4x+14)>6

solve:
(-4x+14)>6
-4x>-8
x<2
(symbol changes when ÷/× by negative for inequalities.)

-(-4x+14)>6
-4x+14<-6
-4x<-20
x>5

the answers are x<2 and x>5, so you would write your answer like this: (-∞,2) ∪ (5,∞)

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A public health organization reports that 40%of baby boys 6-8 months old in the United

Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502$

Hence:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675$

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

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4 0

1 year ago

An 8 oz box of cocoa puffs costs $1.59 and a 2-pound box costs $6.79. which box is a better buy

VikaD [51]

Answer: 8 oz..

1 pound= 16oz if 8oz=1.59 so add it up 4 times 8+8+8+8= 32 which = 2 pounds total 8=1.59 8+8= 1.59x2=3.18 8+8+8=1.59x3=4.77 8+8+8+8/ 1.59x4= 6.36 so all of them eights will = up to 2 pounds and 8 is obviously less

Step-by-step explanation:

6 0

3 years ago

Find ? (inscribed angle)

Lera25 [3.4K]

9514 1404 393

Answer:

107°

Step-by-step explanation:

Opposite angles of an inscribed quadrilateral are supplementary.

X = 180° -Q

X = 180° -73° = 107°

3 0

2 years ago

How do you write 304,800,400 in expanded form and standard

Elanso [62]

Three hundred four million, Eight hundred thousand Four hundred
and standard form is the way you wrote it, 304,800,400

3 0

2 years ago

Which do the following options represent the measure of arc RA

ANTONII [103]

Answer:

77.2°

Step-by-step explanation:

Consider the triangle JKR.

∠KJR=108.6 (lies on the same line as ∠RJA, angles on a straight line add up to 180)

All the angles in a triangle add up to 180, so:

∠JKR+∠KJR+∠JRK=180

∠JKR+108.6+32.8=180

∠JKR=38.6=∠RKA

Consider ∠RKA. This angle stands on the same arc as ∠RCA.

Since the angle at centre is twice the angle at circumference, 2(∠RKA) = ∠RCA.

2(∠RKA) = ∠RCA

2(38.6)=∠RCA

∠RCA=77.2°

5 0

2 years ago