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4 years ago

Please help asap 50 pts

Mathematics

2 answers:

Natasha_Volkova [10]4 years ago

8 0

If you factor the expression by grouping you should get (3g - 5) (2g +7)

Steps:
1. Factor 11 out of 11g
6g^2 + 11 (g) - 35

2. Rewrite 11 as -10 plus 21
6g^2 + (-10 + 21) g - 35

3. Apply the distributive property
6g^2 + (-10g + 21g) - 35

4. Remove parentheses
6g^2 - 10g + 21g - 35

5. Group the first two and the last two terms
(6g^2 - 10g) + (21g - 35)

6. Factor out the greatest common factor from each group
2g (3g - 5) + 7 (3g - 5)

7. Factor out the greatest common factor, 3g - 5
(3g - 5) (2g + 7)

Therefore the answer is A

Allushta [10]4 years ago

7 0

$6g^2+11g-35=6g^2+21g-10g-35=3g(2g+7)-5(2g+7)=(2g+7)(3g-5)$
Answer: A. (3g - 5)(2g + 7)

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3 years ago

In a survey of adults who follow more than one sport, 30% listed football as their favorite sport. You survey 15 adults who foll

maxonik [38]

Answer: 0.5

Step-by-step explanation:

Binomial probability formula :-

$P(x)=^nC_x\ p^x(q)^{n-x}$ , where P(x) is the probability of getting success in x trials , n is the total trials and p is the probability of getting success in each trial.

Given : The probability that the adults follow more than one game = 0.30

Then , q= 1-p = 1-0.30=0.70

The number of adults surveyed : n= 15

Let X be represents the adults who follow more than one sport.

Then , the probability that fewer than 4 of them will say that football is their favorite sport,

$P(X\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\=^{15}C_{0}\ (0.30)^0(0.70)^{15}+^{15}C_{1}\ (0.30)^1(0.70)^{14}+^{15}C_{2}\ (0.30)^2(0.70)^{13}+^{15}C_{3}\ (0.30)^3(0.70)^{12}+^{15}C_{4}\ (0.30)^4(0.70)^{11}\\\\=(0.30)^0(0.70)^{15}+15(0.30)^1(0.70)^{14}+105(0.30)^2(0.70)^{13}+455(0.30)^3(0.70)^{12}+1365(0.30)^4(0.70)^{11}\\\\=0.515491059227\approx0.5$

Hence, the probability rounded to the nearest tenth that fewer than 4 of them will say that football is their favorite sport =0.5

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3 years ago

If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person se

notsponge [240]

Answer:

2.28% probability that a person selected at random will have an IQ of 110 or higher

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 100, \sigma = 5$