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SVETLANKA909090 [29]

2 years ago

6

The formula for electrical power (P, in watts) is where V is the voltage in volts and R is the resistance in ohms. For a particu

lar brand of water heater, R is 32 ohms and P is 0.5 watts. The formula to find V is , and the voltage at which the heater operates is volts.

1 answer:

bazaltina [42]2 years ago

8 0

Answer: 4 V

Step-by-step explanation:

The formula for electrical power $P$ is:

$P=\frac{V^{2}}{R}$

Where:

$P=0.5 W$ is the water heater electrical power

$V$ is he water heater voltage

$R=32 \Omega$ is the electrical resistance

Isolating $V$ :

$V=\sqrt{PR}$

$V=\sqrt{(0.5 W)(32 \Omega)}$

Hence:

$V=4 V$

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nora is planning a birthday party for her little sister, colleen. she needs to purchase 12 cupcakes she can not spend more than

Lapatulllka [165]

Answer:

if she cannot spend more than $40, she will spend .3 on each cupcake

Step-by-step explanation:

when you divide 12 by 40, you get .3 and to check if it is true, you can multiply .3 x 40 and you will get 12 hope you get it right, and have a good day :)

6 0

2 years ago

Which table identifies the one-sided and two-sided limits of function at x = 2?

Yuliya22 [10]

Answer:

Table 3

Step-by-step explanation:

Check table three;

$lim\:_{x\to \:2^-}\:f\left(x\right)=\:4$

$lim\:_{x\to \:2^+}\:f\left(x\right)=\:1$

Since the left hand limit $(lim\:_{x\to \:2^-}\:f\left(x\right))$ is not equal to the right hand limit $(lim\:_{x\to \:2^+}\:f\left(x\right))$ , the limit as x approaches to 2 does not exist.

Therefore "nonexistent" is true, and table 3 is the correct model of the limits of the function at x = 2

6 0

3 years ago

the stemplot below represents the distribution of the test scores for the final exam in a high school algebra course. what was t

Deffense [45]

Answer:

<u><em></em></u>

<u><em>Option A. 94</em></u>

Explanation:

The column of digits to the left (before the vertical line) contains the largest place-value digits.

Here is how you read the values:

First row: 65

Second row: 73, 74, 78

Third row: 80, 80, 81, 82, 85, 88, 89

Fourth row: 92, 93, 94.

Thus, the highest score received is 94.

8 0

2 years ago

Round 32.697 to the nearest tenth hundredth and whole number

zalisa [80]

To the nearest tenth : 32.7

To the nearest hundredth : 32.70

Whole Number : 33

Hope this helps!

5 0

3 years ago

A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0

3 years ago