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Marianna [84]
3 years ago
6

Suppose a number is rounded to the nearest hundred thousand. What is a number greater than 700,000 that rounds to 700,000?

Mathematics
1 answer:
tatiyna3 years ago
5 0
Any number greater than 700,000 with ten-thousands digit 1 through 4 will round down to 700,000.  For example, 748,235 will round to 700,000.
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You cross two true-breeding parents (one with purple flowers and one with white flowers). All of the F1 progeny (offspring) have
kogti [31]

Answer:

The correct answer is 375 purple flowers.

Step-by-step explanation:

If all the F1 generation from true-breeding parents are purple than the genotype would be heterozygous i.e., Pp and Pp. Here P indicates the dominant allele which produces purple flower and p indicates the recessive allele of white flower.

So if we will cross the F 1 generation than the genotype would be PP, Pp, Pp, pp. That means 3 out of 4 offspring would be purple because three genotypes out of four have dominant purple allele(P).

So if out of 4 flower 3 is purple than according to calculation out of 500 flower 375 offspring would be purple(3/4*500= 375). Therefore the correct answer is 375 purple flowers should be produced.

7 0
3 years ago
Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be
hodyreva [135]

Answer:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

And the sample deviation with the following formula:

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

And the sample deviation with the following formula:

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

8 0
3 years ago
Dragon had a lot of bands 12 of them tour then he had 39 left how many did he have at the start
GuDViN [60]

Answer:

51

Step-by-step explanation:

12+39=51

8 0
3 years ago
What is the value of x in the linear equation below?<br><br><br> 4+3x-ax=9-7x+bx
Furkat [3]

Answer:

x = 5/(10 - a - b)

Step-by-step explanation:

Solve for x:

4 + 3 x - a x = 9 - 7 x + b x

Collect in terms of x:

x (3 - a) + 4 = 9 - 7 x + b x

Collect in terms of x:

x (3 - a) + 4 = x (b - 7) + 9

Subtract (b - 7) x + 4 from both sides:

x (10 - a - b) = 5

Divide both sides by 10 - a - b:

Answer:  x = 5/(10 - a - b)

7 0
3 years ago
Solve for r 4/r = 5/7
katen-ka-za [31]

Work is provided in the image attached.

6 0
3 years ago
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