What is 0.800 rounded to nearest tenth

Oliga [24]

For this case we have that the weight of the bouquet of flowers will be:
4.3 * 10 ^ 4 - 2 * (7.5 * 10 ^ 2)
Rewriting we have:
4.1 * 10 ^ 4 miligrams
Thus, we have that the values of p and q are given by:
p = 4.1
q = 4
Answer:
4.1 * 10 ^ 4 miligrams
p = 4.1
q = 4

8 0

3 years ago

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On school days, Kiran walks to school. Here are the lengths of time, in minutes,

Alex

Answer:

14 minutes

Step-by-step explanation:

70 total minutes divided by 5

5 0

2 years ago

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What is the arc measure of BAD on circle P in degrees?

lianna [129]

Answer:

244

Step-by-step explanation:

74+42=116

360-116=244.

5 0

2 years ago

F r e e points if you can answers these questions. Get 4 or the most question correct and ill mark brainliest.

gladu [14]

Answer: 11.Because she could of landed on someting

Step-by-step explanation:

8 0

2 years ago

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Thirty percent of all customers who enter a department store will make a purchase. Suppose that 6 customers enter the store and

torisob [31]

Answer:

Step-by-step explanation:

Let's use binomial distribution, because we are interested in find the total number of successes in a sequence of n independent experiments.

In probability, a binomial distribution is used in a Bernoulli process. That is, a sequence of experiments of n independent trials, each asking a dichotomous question, that means it only has two answers. One answer is a success (probability: p) and the other is a failure (probability: 1-p). In this case, every customer who enters in the store is the experiment. If they make a purchase is a success event, taking into account that every purchase decision has to be independent.

$P(X=x) = C(n,x)p^{x} (1-p)^{n-x}$

x= total number of success

n= total number of experiments

p=probability of success on an indivual trial

$C(n,x) = \frac{n!}{x!(n-x)!}$

a) x= 5 ------> customers that make a purchase

n= 6 ------> total customers

p=0.3

$p(X=5)= C(6,5)(0.3)^{5} (1-0.3)^{6-5} \\p(X=5)=0.102$

b) At least 3 customers make a purchase. x ≥ 3

$P(X\geq 3)=1-P (X=0)-P(X=1)-P(X=2)\\P(X\geq 3)=1-(C(6,0)(0.3^{0})(0.7)^{6}) - (C(6,1)(0.3^{1})(0.7)^{5}) - (C(6,2)(0.3^{2})(0.7)^{4})\\ P(X\geq 3) = 1-0.11765-0.30253-0.32414\\P(X\geq3)= 0.2557$

c) At most 2 customers make a purchase. x ≤ 2

$P(X\leq2) = P(X=0) + P(X=1) + P(X=2)\\P(X\leq2) = (C(6,0)(0.3^{0})(0.7)^{6}) + (C(6,1)(0.3^{1})(0.7)^{5}) + (C(6,2)(0.3^{2})(0.7)^{4})\\P(X\leq2) = 0.11765-0.30253-0.32414\\P(X\leq2)=0.7443$

d)At least 1 customer makes a purchase. x ≥ 1

$P(X\geq 1)=1-P(X=0)\\P(X\geq 1)= 1-C(0,6)(0.3)^{0}(0.7)^{6} \\P(X\geq 1)= 1-0.11765 = 0.8824$

e)They must be kind to customers, always willing to help and give information about products clearly and honestly.

Have enough inventory and offer promotions for the purchase of more than one product

Have an agile payment system that avoids rows and delays in purchases.

6 0

3 years ago