PLEASE HELP ASAP 25 PTS

What is 120.571 in expanded form?

borishaifa [10]

The expanded form would be 100.000 +20.000+0.000+.500+.070+.001

5 0

2 years ago

Read 2 more answers

13. For what value of b would the following system of equations have an infinite number of solutions?

tiny-mole [99]

Answer:

b=2

Step-by-step explanation:

we have

9x+12y=21 -----> equation A

6x+8y=7b ----> equation B

we know that

If the system of equations have an infinite number of solutions then the equation A must be equal to the equation B

Multiply equation B by 1.5 both sides

1.5*[6x+8y[=7b*1.5

9x+12y=10.5b ----> equation C

Compare equation A and equation C

9x+12y=21 -----> equation A

9x+12y=10.5b ----> equation C

For the equations to be equal it must be fulfilled that

21=10.5b

solve for b

b=21/10.5

b=2

5 0

2 years ago

If a randomizer picks seven letters from the alphabet (with repeats allowed), how many possible outcomes are there?

Virty [35]

Answer:

7×26=182

Step-by-step explanation:

I'm pretty sure that's the right answer but I just multiplied 26 by 7. Hope that helped!

7 0

2 years ago

Select all the true statements. *

I am Lyosha [343]

A,C,D are the answers

8 0

2 years ago

A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio

Reptile [31]

Answer:

The probability is $P(|p-\^{p}| > 0.03) = 0.0040$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.07$

The mean of the sampling distribution is $\mu_p = 0.07$

The sample size is n = 600

Generally the standard deviation is mathematically represented as

$\sigma_p = \sqrt{\frac{p (1 -p)}{n} }$

=> $\sigma_p = \sqrt{\frac{0.07(1 -0.07)}{600} }$

=> $\sigma_p = 0.010416$

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

$P(|p-\^{p}| > 0.03) = 1 - P(|p -\^{p}| \le 0.03)$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-0.03 \le p -\^{p} \le 0.03 )$

Now add p to both side of the inequality

=> $P(|p-\^{p}| > 0.03) = 1 - P( 0.07-0.03 \le \^{p} \le 0.03+ 0.07 )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(0.04 \le \^{p} \le 0.10 )$

Now converting the probabilities to their respective standardized score

=> $P(|p-\^{p}| > 0.03) = 1 - P(\frac{0.04 - 0.07}{0.010416} \le Z \le \frac{0.10 -0.07}{0.010416} )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-2.88 \le Z \le 2.88 )$

=> $P(|p-\^{p}| > 0.03) = 1 - [P(Z \le 2.88) - P(Z \le -2.88)]$

From the z-table

$P(Z \le 2.88) = 0.9980$

and

$P(Z \le -2.88) = 0.0020$

So

$P(|p-\^{p}| > 0.03) = 1 - [0.9980 - 0.0020]$

=> $P(|p-\^{p}| > 0.03) = 0.0040$

7 0

2 years ago