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3 years ago

What will most likely happen to a characteristic that has a high survival value in a population?

Mathematics

1 answer:

Levart [38]3 years ago

6 0

The characteristic will spread and be more common throughout populations.

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One serving of chad's cereal is 1 1/4 ounces. How many servins are in 15 1/3 ounces box?

IrinaK [193]

change the mixed numbers to improper fractions before dividing

1 $\frac{1}{4}$ = $\frac{5}{4}$ and 15 $\frac{1}{3}$ = $\frac{46}{3}$

divide contents of box by a serving

$\frac{46}{3}$ ÷ $\frac{5}{4}$

change ÷ to × and turn the second fraction upside down

= $\frac{46}{3}$ × $\frac{4}{5}$

= $\frac{184}{15}$ = 12.26666 = 12 complete servings

5 0

3 years ago

Yanni threw his paper airplane 15 1/2 feet. Adrian threw his paper airplane 3/4 of yanni's distance. What is the distance Adrian

vagabundo [1.1K]

Answer:

19.125

Step-by-step explanation:

multiply 15.5 by 3 and then divide the answer by 4

4 0

3 years ago

Which of these is an open sentence?

Anna007 [38]

B because k is not a specified number

3 0

3 years ago

Read 2 more answers

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

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

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]