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3 years ago

(4.6•10^9)+ (1.7•10^7)

Mathematics

1 answer:

Gennadij [26K]3 years ago

6 0

Answer:

4617000000

Step-by-step explanation:

10^9 = 1000000000

1000000000 x 4.6 = 4600000000

10^7 = 10000000

10000000 x 1.7 = 17000000

4600000000 + 17000000 = 4617000000

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CAN SOMEONE HELP WITH THIS ASAP!!!

jek_recluse [69]

Answer: x=8

Step-by-step explanation:

Since we are given that ∠ABC and ∠DBC are complementary angles, we know that adding them together should give us 90°. We can use this to solve for x.

6x+13+4x-3=90 [combine like terms]

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10x=80 [divide both sides by 10]

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Now, we know that x=8.

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4 years ago

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sukhopar [10]

Answer:

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Step-by-step explanation:

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4 years ago

What is the coefficient of the term 12p12p12, p in the expression 12p+ 9q12p+9q12, p, plus, 9, q?

klio [65]

Answer:

the coefficient is 12

Step-by-step:a numerical or constant quantity placed before and multiplying the variable in an algebraic expression explanation:

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3 years ago

Measure the line segments and create a line plot with the data

Fantom [35]

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3 years ago

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

3 years ago