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2 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz help

Mathematics

2 answers:

KiRa [710]2 years ago

4 0

Answer:

Juan sent 34 text messages.

Step-by-step explanation:

$52 - $35 = $17

$0.50 * 34 = 17

$35 + $17 = $52

Juan the horse.

salantis [7]2 years ago

3 0

Answer:

Step-by-step explanation:

52-35=17

17/0.5=34

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Olivia used her past health history and information about her doctor visits to create this table to compare health costs with an

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The expected value of health care without insurance is $437.25.

The expected value of health care with insurance is $1,636.40.

<h3>What are the expected values?</h3>

The expected values can be determined by multiplying the respective probabilities by its associated costs.

The expected value of health care without insurance

= (1 x 0) + (0.32 x 1050) + (0.45 x $225)

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The expected value of health care with insurance

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Thus, The expected value of health care without insurance is $437.25.

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3 years ago

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3 years ago

A random sample of 9 fields of corn has a mean yield of 40.6 bushels per acre and standard deviation of 7.52 bushels per acre. D

zaharov [31]

Answer:

The critical value that should be used in constructing the confidence interval is -1.397 and 1.397.

80% confidence interval for the true mean yield is [37.1 bushels per acre, 44.1 bushels per acre].

Step-by-step explanation:

We are given that a sample of 1584 third graders, the mean words per minute read was 35.7. Assume a population standard deviation of 3.3.

Firstly, the pivotal quantity for 80% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample mean yield = 40.6 bushels per acre

s = sample standard deviation = 7.52 bushels per acre

n = sample of fields of corn = 9

$\mu$ = true mean yield

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about the population standard deviation.

So, 99% confidence interval for the true mean yield, $\mu$ is ;

P(-1.397 < $t_8$ < 1.397) = 0.80 {As the critical value of t at 8 degree of

of freedom are -1.397 & 1.397 with P = 10%}

P(-1.397 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.397) = 0.80

P( $-1.397 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.397 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.397 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.397 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.397 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.397 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $40.6-1.397 \times {\frac{7.52}{\sqrt{9} } }$ , $40.6+1.397 \times {\frac{7.52}{\sqrt{9} } }$ ]

= [37.1 , 44.1]

Therefore, 80% confidence interval for the true mean yield is [37.1 bushels per acre, 44.1 bushels per acre].

7 0

3 years ago