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Phantasy [73]
3 years ago
12

Mr grant needs 30 pieces of felt he can buy them in packs of 10 or as single pieces what are the different ways mr

Mathematics
1 answer:
Olin [163]3 years ago
6 0

The <em><u>correct answer</u></em> is:

He can buy 3 packs of 10 or 30 single pieces.

Explanation:

He needs 30 pieces.  Since they are sold in 10 packs, it would take 30/10 = 3 packs of 10.

Other than this, the pieces are sold as singles.  This means he could also buy 30 single pieces.

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Tax is paid on meals in a restaurant.
liberstina [14]

Answer:

17.52%

Step-by-step explanation:

65.35/11.45 = 0.1752

.1752 * 100

= 17.52

4 0
2 years ago
7.5% of the grade has green eyes if there are 120 students in the grade how many dont have green eyes​
GREYUIT [131]

Answer:

Step-by-step explanation:

If 7.5 % has green eyes, then 92.5% (the complement given by 100%-7.5%)  dont have green eyes. 92.5% of 120 is given by

0.925*(120)=111

111 students in the grade dont have green eyes.  

8 0
3 years ago
On Sunday, chara ran 6 more miles than he did on Saturday. If he ran m miles on Saturday, which expression represents how many m
sattari [20]

Answer:

Is A supposed to be 6 x m or no??

Step-by-step explanation:

8 0
3 years ago
A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain
svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55

The mean of this distribution is:

\mu=np=10*0.6=6

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8

The probability of having between 4 and 8 customers choosing the oversize version is:

P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328

7 0
3 years ago
Rebecca draws a graph of a real world relationship that turns out to be a set of unconnected points . Can the relationship be li
Bumek [7]
Well, A linear relationship has to be a straight line and a proportional relationship has to go through the origin! Hope that's a good hint!
8 0
3 years ago
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