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3 years ago

What is the product written in scientific notation? (5.91×10−3)⋅(8.7×1010)

2 answers:

5 0

$(5.91\cdot10^{-3})\cdot(8.7\cdot10^{10})=(5.91\cdot8.7)\cdot(10^{-3}\cdot10^{10})\\\\=51.417\cdot10^{-3+10}=51.417\cdot10^7=5.1417\cdot10\cdot10^7\\\\=\boxed{5.1417\cdot10^8}\to\boxed{B.}$

$Used:\\\text{Associative property}\ (a\cdot b)\cdot c=a\cdot(b\cdot d)\\\\a^n\cdot a^m=a^{n+m}$

romanna [79]3 years ago

4 0

(5.91×10^−3)⋅(8.7×10^10)

= (5.91 x 8.7) x 10^(- 3 + 10)

= 51.417 x 10^7

= 5.1417 x 10^8

Answer is

B. 5.1417 × 10^8

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In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. This is the 14

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We are confident at 99% that the difference between the two proportions is between $0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420$

Step-by-step explanation:

Part a

Data given and notation

$X_{D}=3266$ represent the number people registered as Democrats

$X_{R}=2137$ represent the number of people registered as Republicans

$n=7525$ sampleselcted

$\hat p_{D}=\frac{3266}{7525}=0.434$ represent the proportion of people registered as Democrats

$\hat p_{R}=\frac{2137}{7525}=0.284$ represent the proportion of people registered as Republicans

The standard error is given by this formula:

$SE=\sqrt{\frac{\hat p_D (1-\hat p_D)}{n_{D}}+\frac{\hat p_R (1-\hat p_R)}{n_{R}}}$

And the standard error estimated given by the problem is 0.008

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion of Democrats that approve of the way the California Legislature is handling its job

$\hat p_A =\frac{1894}{3266}=0.580$ represent the estimated proportion of Democrats that approve of the way the California Legislature is handling its job

$n_A=3266$ is the sample size for Democrats

$p_B$ represent the real population proportion of Republicans that approve of the way the California Legislature is handling its job

$\hat p_B =\frac{385}{2137}=0.180$ represent the estimated proportion of Republicans that approve of the way the California Legislature is handling its job

$n_B=2137$ is the sample for Republicans

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.580-0.180) - 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.380$

$(0.580-0.180) + 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.420$

And the 99% confidence interval would be given (0.380;0.420).

We are confident at 99% that the difference between the two proportions is between $0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420$

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