Please help me now plz I will mark you brainliest

Thaddeus shakes his head as he pays $12.37 to the clerk for lunch. Complaining to a friend, Thaddeus says, “Do you realize that

atroni [7]

a)

his income must be "x" which is the 100%, and we would know that 0.2% of that is 12.37.

$\begin{array}{ccll} amount&\%\\ \cline{1-2} x&100\\ 12.37&0.2 \end{array}\implies \cfrac{x}{12.37}=\cfrac{100}{0.2}\implies 0.2x=1237 \\\\\\ x=\cfrac{1237}{0.2}\implies x=61.85$

b)

his income must be "x" which is the 100%, and we would know that 2% of that is 12.37.

$\begin{array}{ccll} amount&\%\\ \cline{1-2} x&100\\ 12.37&2 \end{array}\implies \cfrac{x}{12.37}=\cfrac{100}{2}\implies 2x=1237 \\\\\\ x=\cfrac{1237}{2}\implies x=618.5$

btw he's not making much either way.

7 0

1 year ago

Lines j and k are parallel how to solve

Ede4ka [16]

I can't answer your question without a picture... Sorry.

7 0

2 years ago

The lowest temperature ever recorded at Oymyakon in Russia was 96.2°F below 0°F. The lowest temperature ever recorded at Prospec

Alekssandra [29.7K]

Answer:

Thermometer reading of the lowest recorded temperature at Oymyakon was -96.2° F

Thermometer reading of the lowest recorded temperature at Prospect Creek was -80° F

Step-by-step explanation:

If the temperature is x° F below 0° F then the thermometer reading is -x° F

It is given that the Lowest temperature recorded at Oymyakon in Russai was 96.2°F below 0°F

So the thermometer reading of the lowest recorded temperature at Oymyakon was -96.2° F

Also it is given that the Lowest temperature recorded at Prospect Creek in Alaska was 80°F below 0° F

So the thermometer reading of the lowest recorded temperature at Prospect Creek was -80° F

7 0

3 years ago

"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen

Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± $Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ ]