$1)\quad \Re(9+3i)=9\qquad\text{F}\\\\
2)\quad \Im(9+3i)=3\qquad\text{E}\\\\
3)\quad \overline{(9+3i)}=9-3i\qquad\text{D}\\\\
4)\quad 3i-(9+3i)=3i-9-3i=-9\qquad\text{A}\\\\
5)\quad 9+3i-9=3i\qquad\text{B}\\\\
6)\quad 9-(9+3i)=9-9-3i=-3i\qquad\text{C}\\\\$

5 0

3 years ago

What is the sum of 2.5 × 104 and 3.4 × 104? (1 point)

Tju [1.3M]

Answer:

5.9× 10^4

Step-by-step explanation:

I'm assuming that by "104" you actually meant "10^4," which reads "fourth power of 10." Please be sure to use the symbol " ^ " to express exponentiation.

The sum of 2.5 × 10^4 and 3.4 × 10^4 is (2.5 + 3.4)×10^4

5.9× 10^4

7 0

3 years ago

The most recent public health statistics available indicate that 23.6% of American adults smoke cigarettes. Using the 68-95-99

artcher [175]

Answer:

There is a 68% chance that between 17% and 30% are smokers.

There is a 95% chance that between 10% and 37% are smokers.

There is a 99.7% chance that between 4% and 44% are smokers.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of the sampling distribution of sample proportion is:

$\mu_{\hat p}=p\\$

The standard deviation of the sampling distribution of sample proportion is:

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

Given:

n = 40

p = 0.236

Compute the mean and standard deviation of this sampling distribution of sample proportion as follows:

$\mu_{\hat p}=p=0.236$

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.236(1-0.236)}{40}}=0.067$

The Empirical Rule states that in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be divided into three parts:

68% data falls within 1 standard-deviation of the mean.

That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.

95% data falls within 2 standard-deviations of the mean.

That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.

99.7% data falls within 3 standard-deviations of the mean.

That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

Compute the range of values that has a probability of 68% as follows:

$P (\mu_{\hat p} - \sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + \sigma_{\hat p}) = 0.68\\P(0.236-0.067\leq \hat p \leq 0.236+0.067)=0.68\\P(0.169\leq \hat p \leq0.303)=0.68\\P(0.17\leq \hat p \leq0.30)=0.68$

Thus, there is a 68% chance that between 17% and 30% are smokers.

Compute the range of values that has a probability of 95% as follows:

$P (\mu_{\hat p} - 2\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 2\sigma_{\hat p}) = 0.95\\P(0.236-2\times 0.067\leq \hat p \leq 0.236+2\times0.067)=0.95\\P(0.102\leq \hat p \leq 0.370)=0.95\\P(0.10\leq \hat p \leq0.37)=0.95$

Thus, there is a 95% chance that between 10% and 37% are smokers.

Compute the range of values that has a probability of 99.7% as follows:

$P (\mu_{\hat p} - 3\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 3\sigma_{\hat p}) = 0.997\\P(0.236-3\times 0.067\leq \hat p \leq 0.236+3\times0.067)=0.997\\P(0.035\leq \hat p \leq 0.437)=0.997\\P(0.04\leq \hat p \leq0.44)=0.997$

Thus, there is a 99.7% chance that between 4% and 44% are smokers.

7 0

2 years ago