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2 years ago

If 5.3 pounds of nails cost $16.96 what is the price of one pound of nails?

2 answers:

8 0

This is the word formula we can use to solve this problem:

Price of 1 pound of nails = price for ____# of pounds of nails ÷ ___# of pounds of nails

Our number to fill in the ____ is 5.3.

price for 5.3 pounds of nails: $16.96

This means:
price for 1 pound of nails = $16.96 ÷ 5.3

16.96 ÷ 5.3 = 3.2

Therefore 1 pound of nails costs $3.2.

Rom4ik [11]2 years ago

5 0

$16.96÷5.3=3.2(naila)-price

You might be interested in

What is the value of x?<br><br> 2x + (5x+6)

grin007 [14]

Answer:

The value is 7x + 6

Step-by-step explanation:

8 0

2 years ago

There are three boxes: one with two golden coins, one with two silver coins, and one with one golden coin and one silver coin. A

krek1111 [17]

Answer:

Step-by-step explanation:

From Bayes' theorem is stated mathematically as the following equation:[2]

{\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}

where A and B are events and P(B) ≠ 0.

P(A) and P(B) are the probabilities of observing A and B without regard to each other.

P(A | B), a conditional probability, is the probability of observing event A given that B is true.

P(B | A) is the probability of observing event B given that A is true.

At this point, go through the attached file before you continue with part B.

Part B)

P(silver) = P(silver from SS)+P(silver from GS)

note P(SS)=P(GG)=P(GS) = 1/3

P(silver from SS) = 1

P(silver from GS) = 1/2

hence

P(Silver from SS) = 1/3

P(Silver from GS) = 1/3 *1/2

P(Silver) = 1/3*1+1/3*1/2

required probability = P(Silver from SS)/P(Silver) = 2/3

4 0

2 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

2 years ago

Can any kind soul help me ASAP!

CaHeK987 [17]

Answer:

Hope it helps u

Step-by-step explanation:

As we know that ,

Mean = sum of the terms/ numbers of terms

But here grouped data is given so , we use the formula

Mean=∑[f. m]/ ∑f

where f is frequency and m is mid point of each height ,

Now first we have to find the mid point of each interval, where

midpoint of each interval = (lower boundary + upper boundary)/2

m1=(150+154)/2 = 152

m2=(155+159)/2= 157,now found other by same formula, for each interval

m3= 162

m4= 167

m5=172 Now we find the midpoint of each interval ,so now

∑[f. m]=f1*m1+f2*m2+f3*m3+f4*m4+f5*m5

now putting the values of each frequency for given interval and midpoint of each interval we will get,

∑[f. m]=456+942+1296+167*x+344 = 167*x+3038

Now find,

∑f=f1+f2+f3+f4+f5

∑f=19+x

Now we have,

∑[f. m]=167*x+3038

∑f=19+x

also given mean height=161.6 cm

putt these values in above equation we get,

161.6= $\frac{167*x+3038}{19+x}$

now solve this ,

161.6(19+x)=167*x+3038

3070.4+161.6*x=167*x+3038

3070.4-3038=167*x-161.6*x

32.4=5.4*x

x=32.4/5.4

3 0

2 years ago

Joseph wants to estimate 1 3/7 . 5 4/5 To estimate, he simplifies the expression 1 1/2 . 6 Is Joseph’s estimate an underestimate

Zolol [24]

Well if her simplifies 1 3/7 to 1 1/2 and 5 4/5 to 6 then he would be overestimating because 1 1/2 is greater than 1 3/7 and 6 is greater than 5 4/5

Hope this helps!

7 0

2 years ago