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3 years ago

What is 123.210,000 in word form

Mathematics

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:

123.210,000

Step-by-step explanation:

one hundred and twenty-three million two hundred ten thousand

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Can y’all plz help and thank you. Aslo hope y’all have a wonderful and beautiful day.

stepan [7]

Answer:

a) Rumble: $1

Pedal: $0.6

b) more than 5 songs

c) No, should be integers only

d) No, has to be integers greater than or equal to 2

Step-by-step explanation:

Part a

Rate per song is the slope.

Rumble:

Slope is 0.6

Rate is $0.6 per song

Pedal:

Slope: (4-2)/(2-0) = 1

Rate is $1 per song

Part b

Pedal would be a better deal if it's cost is lower than Rumble's

Rumble's cost: C = 2 + S

Pedal's cost: C = 4 + 0.6S

4 + 0.6S < 2 + S

0.4S > 2

S > 5

c) No, because no. of songs has to be positive integers only.

For example graph shows S = 2.5, but we know that's not possible

d) No, since the cost is 2 + S,

cost can't be in decimals either.

Cost too should be only integers greater than equal to 2

7 0

3 years ago

Which of these is a trinomial?

7nadin3 [17]

I’m thinking it’s D but i’m not too sure

8 0

3 years ago

Read 2 more answers

Natalie's family drank 0.9 of the 2.6 liters of the lemonade she made for a picnic. how much lemonade did her family drink? Show

rewona [7]

Answer:

There is 1.7 left

Step-by-step explanation:

2.6 - 0.9 = 1.7

5 0

3 years ago

Whats 3a+a+5b-2b using the indicated operations

Pepsi [2]

Answer:

4a + 3b

Step-by-step explanation:

3a + a + 5b - 2b

3a + s = 4a

5b - 2b = 3b

6 0

3 years ago

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Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

3 years ago