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2 years ago

Given below is the odd-number function. which of the following are equal to o(7)?

Mathematics

2 answers:

Sav [38]2 years ago

4 0

If this is a multiple choice question, try to see if -7 and 7 come up as a pair or just -7. I think it is -7.

sveta [45]2 years ago

4 0

We have to see the multiple-choice answers.

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the price of a bike at store A is 5/6 the price at store B . The price at store A is 150.60.wriite an equation and solve to see

Tasya [4]

(5/6) 100
------ = ------
x 150.60

7 0

2 years ago

Is the negative square root 400 rational or irrational

Aleks [24]

Answer:

Step-by-step explanation:

Rational

8 0

2 years ago

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At Chrissy's Clothes Warehouse, it takes fraction 2 over 5 of a day to complete fraction 1 over 10 of an order of t-shirts. At t

erastovalidia [21]

1/10 of all take 2/5 time
10/10=10 times 1/10
so 10 times 2/5 time=20/5=4 days to finish 1 order

4 days

4 0

2 years ago

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Busi has a bag containing 20 balls. There are twice as much yellow balls than blue balls, but yellow balls are only a third of r

Darina [25.2K]

Answer:

12 red balls

4 yellow balls

2 blue balls

2 green balls

Step-by-step explanation:

set no. of yellow, blue, red and green ball as y, b, r, g respectively

y=2b

r=3y

g=b

y+b+r+g = 20 (translate all of them into b (smallest in qty)

2b + b + 3(2b) + b = 20

10b = 20

b =2

y = 4, r = 12, b =2, g =2

5 0

1 year ago

A genetic experiment involving peas yielded one sample of offspring consisting of 430 green peas and 155 yellow peas. Use a 0.05

Dimas [21]

Answer:

the null and alternative hypothesis are

$H_{0}$ : p=0.24

$H_{a}$ : p≠0.24

Test statistic is 1.416.

P-Value is 0.1568

We fail to reject the null hypothesis under 0.05 significance level

<em>At 0.05 significance</em> we cannot say that the proportion of offspring peas will be yellow is not 0.24 (24%)

Step-by-step explanation:

Let p be the proportion of offspring peas will be yellow. Then,

$H_{0}$ : p=0.24

$H_{a}$ : p≠0.24

test statistic can be calculated as

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of yellow offspring peas ( $\frac{155}{155+430} =0.265$
p is the proportion of yellow offspring peas assumed under null hypothesis. (0.24)

N is the sample size (585)

Test statistic is z= $\frac{0.265-0.24}{\sqrt{\frac{0.24*0.76}{585} } }$ ≈ 1.416.

Using z-table, we can find that P-Value is 0.1568

We fail to reject the null hypothesis under 0.05 significance level since 0.157>0.05

We can conclude that <em>at 0.05 significance</em> we cannot say that the proportion of offspring peas will be yellow is not 0.24 (24%)

6 0

3 years ago