Whats -0.07776 as a fraction?

Assuming A = dominant allele that produces red-eye, a = recessive allele that produces sepia eye, B = dominant allele that produ

AURORKA [14]

Answer:

25% of probabilities

Step-by-step explanation:

A= red eye

a=sepia eye

B=longwing

b=apterous

AB aB

Ab AABb AaBb

ab AaBb aaBb

AABb=25%

AaBb=25%

Ab= 25%

aaBb=25%

aaBb= sepia eye with longwig

4 0

3 years ago

Help plsssssssssss i really need help

mariarad [96]

Answer:

A= 256 stickers B= 8 stickers per student C= 1 more box

Step-by-step explanation:

3 0

2 years ago

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Precipitation for the month of March was 6 1/2 inches less than average. In April, the precipitation was 2 3/8 inches less than

lidiya [134]

Answer:

See explanation.

Step-by-step explanation:

It is subtraction

whole numbers, then numerators.

6 - 2 = 4

1/2 - 3/8

= 4/8 - 3/8 (1/2 = 4/8)

= 1/8

= 4 1/8

hope this helps!

8 0

2 years ago

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Explain how you can tell that 31/33 is in simplest form

shtirl [24]

Well 31 is not able to go any were and it has to stay 31

3 0

2 years ago

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Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0

3 years ago