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2 years ago

I need help with these 2

Mathematics

1 answer:

VashaNatasha [74]2 years ago

5 0

For the first one you need to do
x+2+x+2+2x+4+2x+4
=6x+12

When doing this kind of problem you should use FOIL
First, Outer, Inner, Last
(x+2)(2x+4)
First- x•2x= 2x^2
Outer- x•4= 4x
Inner- 2•2x= 4x
Last- 2•4= 8

Combined you have
=2x^2+8x+8

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The number of players in a tennis tournament is reduced by half (50%) after every round. In round 1 there are 256 player, in rou

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Answer:

$a_{n} = a_{1} (\frac{1}{2} )^{n - 1}$ where $a_{1} = 256$ .

The number of players compete in round 7 will be 4.

Step-by-step explanation:

The number of players in a tennis tournament is reduced by half (50%) after every round. In round 1 there are 256 players, in round 2 there are 128 players, and so on.

So, the number of players is multiplied by $\frac{1}{2}$ gives the number of players in the next round.

Therefore, the explicit function of number of players in nth round is given by

$a_{n} = a_{1} (\frac{1}{2} )^{n - 1}$ where $a_{1} = 256$ .

Therefore, the number of players compete in round 7 will be $a_{7} = 256 \times (\frac{1}{2} )^{(7 - 1)} = 4$ . (Answer)

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3 years ago

What is the greatest common factor (GCF) of 40 and 42?

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Answer: GCF of 40 and 42 is 2

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3 years ago

Divide. Write your answer as a fraction in simplest form. − 7 1/0÷2/5=

forsale [732]

Answer:

$- \frac{71}{4}$

Step-by-step explanation:

$- 7 \frac{1}{10} \div \frac{2}{5}$

$= - \frac{71}{10} \times \frac{5}{2}$

$= - \frac{71}{2} \times \frac{1}{2}$

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3 years ago

) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa

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Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

Let $p_1$ = population proportion of drivers 16-24 years old who buckle up .

$p_2$ = population proportion of drivers 25-69 years old who buckle up .

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no significant difference in seat belt use}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a difference in seat belt use}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers 16-24 years old who buckle up = 79%

$\hat p_2$ = sample proportion of drivers 25-69 years old who buckle up = $\frac{924}{1100}$ = 84%

$n_1$ = sample of 16-24 years old drivers = 1000

$n_2$ = sample of 25-69 years old drivers = 1100

So, test statistics = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a difference in seat belt use.

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