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tatiyna
3 years ago
6

Trail mix is sold in 1 - pound bags.Mary will buy some trail mix and re-package it so that each of the 15 members of her hiking

club gets one 2/5-pound bag.How many 1- pound bags of trail mix should Mary buy to have enough trail mix without leftovers
Mathematics
2 answers:
quester [9]3 years ago
8 0
6 pounds.2/5×15=30/5divide 30 by 5 =6
joja [24]3 years ago
7 0
1 person = 2/5
4/5 2/5 4/5 2/5 4/5 2/5 4/5 2/5 4/5 2/5
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Which is equivalent to sqrt x^10<br> A. X-5<br> B. X -1/5<br> C. X sqrt 10<br> D. X 1/5<br> E. X^5
Lesechka [4]
<h3>Answer:  E) x^5</h3>

\sqrt{x^{10}} = x^5

=====================================================

Explanation:

We simply take half of the exponent 10 to get 5. This applies to square roots only.

So the rule is \sqrt{a^b} = a^{b/2}

A more general rule is

\sqrt[n]{a^b} = a^{b/n}

If n = 2, then we're dealing with square roots like with this problem. In this case, a = x and b = 10.

5 0
3 years ago
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PLEASE HELP!!! I ONLY HAVE A FEW MINUTES!!
TEA [102]

Answer:

y=0.25x-6

Step-by-step explanation:

6 0
3 years ago
maria threw a party for the promotion she received the party cost $20 for two people if she invited 20 guest what would be the c
zhenek [66]

The total cost would  be:

$20 = two people- 2*20=40$

20 = guest

20*20=

400 would be the cost

6 0
3 years ago
Some college graduates employed full-time work more than hours per week, and some work fewer than hours per week. We suspect tha
Alex_Xolod [135]

Answer:

a)Null hypothesis:\mu =40      

Alternative hypothesis:\mu \neq 40      

b) A Type of error I is reject the hypothesis that \mu is equal to 40 when is fact \mu is equal to 40c) We can commit a Type II Error, since by definition "A type II error is the non-rejection of a false null hypothesis and is known as "false negative" conclusion"Step-by-step explanation:Assuming this problem: "Some college graduates employed full-time work more than 40 hours per week, and some work fewer than 40 hours per week. We suspect that the mean number of hours worked per week by college graduates, [tex]\mu , is different from 40 hours and wish to do a statistical test. We select a random sample of college graduates employed full-time and find that the mean of the sample is 43 hours and that the standard deviation is 4 hours. Based on this information, answer the questions below"

Data given

\bar X=43 represent the sample mean

\mu population mean (variable of interest)

s=4 represent the sample standard deviation

n represent the sample size  

Part a: System of hypothesis

We need to conduct a hypothesis in order to determine if actual mean is different from 40 , the system of hypothesis would be:    

Null hypothesis:\mu =40      

Alternative hypothesis:\mu \neq 40      

Part b

In th context of this tes, what is a Type I error?

A Type of error I is reject the hypothesis that \mu is equal to 40 when is fact [tex]\mu is equal to 40

Part c

Suppose that we decide not to reject the null hypothesis. What sort of error might we be making.

We can commit a Type II Error, since by definition "A type II error is the non-rejection of a false null hypothesis and is known as "false negative" conclusion"

5 0
3 years ago
Read 2 more answers
Can someone please help me​
astra-53 [7]

Answer:

square 20 has 44 green squares

square 21 has 45 green squares

Step-by-step explanation:

To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).

For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

L = side length (3,4,5...)

G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2   ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1       now substitute L=n+2

G1(n) = 2(n+2) -1    simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

5 0
3 years ago
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