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2 years ago

A ship is carrying 10,250 pounds of individually boxed TV sets. If each box weighs 20.5 pounds, how many boxes are on the ship?

Mathematics

2 answers:

joja [24]2 years ago

5 0

divide total weight by weight of each box

10250 / 20.5 = 500 boxes total

ololo11 [35]2 years ago

3 0

There are 500 boxes on the ship

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The graph shows the function f(x) = (2.5)x was horizontally translated left by a value of h to get the function g(x) = (2.5)x–h.

ziro4ka [17]

Answer:

-2

Step-by-step explanation:

8 0

1 year ago

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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

2 years ago

Given two vectors a⃗ =4.00i^+7.00j^ and b⃗ =5.00i^−2.00j^ , find the vector product a⃗ ×b⃗ (expressed in unit vectors). what is

FinnZ [79.3K]

The vector product of $\boxed{a \times b = - 43\hat k}$ and the magnitude of $a \times b$ is $\boxed{43}.$

Further explanation:

Given:

Vector a is $\vec a = 4.00\hat i + 7.00\hat j.$

Vector b is $\vec b = 5.00\hat i - 2.00\hat j.$

Explanation:

The cross product of a \times b can be obtained as follows,

$\begin{aligned}a \times b &= \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k} \\4&7&0\\5&{ - 2}&0 \end{array}}\right|\\&= \hat i\left( {0 - 0} \right) - \hat j\left( {0 - 0} \right) + \hat k\left( { - 9 - 35} \right)\\&= 0\hat i - 0\hat j - 43\hat k\\&= - 43\hat k\\\end{aligned}$

The vector can be expressed as follows,

$a \times b = - 43\hat k$

The magnitude of $a \times b$ can be obtained as follows,

$\begin{aligned}\left| {a \times b} \right| &= \sqrt {{0^2} + {0^2} + {{\left( { - 43} \right)}^2}}\\&= \sqrt {{{43}^2}}\\&= 43\\\end{aligned}$ 43404

The vector product of $\boxed{a \times b = - 43\hat k}$ and the magnitude of $a \times b$ is $\boxed{43}.$

Learn more:

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Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Vectors

Keywords: two vectors, vector product, expressed in unit vectors, magnitude, vector a, vector b, a=4.00i^+7.00j^, b=5.00i^-2.00j^, unit vectors, vector space.

8 0

2 years ago

Read 2 more answers

A carpool service has 2,000 daily riders. A one-way ticket costs $5.00. The service estimates that for each $1.00 increase to th

solmaris [256]

Answer:

Total number of riders that ride on carpool daily = 2000

Total Cost of one way ticket = $ 5.00

Total Amount earned if 2000 passengers rides daily on carpool = 2000 × 5

= $10,000

If fare increases by $ 1.00

New fare = $5 + $1

= $6

Number of passengers riding on carpool = 2,000 - 100 = 1,900

If 1,900 passengers rides on carpool daily , total amount earned ,if cost of each ticket is $ 6 = 1900 × $6 = $11400

As we have to find the inequality which represents the values of x that would allow the carpool service to have revenue of at least $12,000.

For $ 1 increase in fare = (2,000 - 1 × 100) passengers

For $ x increase in fare, number of passengers = 2,000 - 100·x

= (2,000 - 100·x) passengers

New fare = 5 + x

New Fare × Final Number of passengers ≥ 12,000

(5+x)·(2,000 - 100 x) ≥ 12,000

5 (2,000 - 100 x) + x(2,000 - 100 x) ≥ 12,000

10,000 - 500 x + 2,000 x - 100 x² ≥ 12,000

100 - 5 x + 20 x - x² ≥ 120

- x² + 15 x +100 - 120 ≥ 0

-x² + 15 x -20 ≥ 0

x² - 15 x + 20 ≤ 0

⇒ x = 1.495

x ≥ $ 1.495, that is if we increase the fare by this amount or more than this the revenue will be at least 12,000 or more .

Also, f'(x) = 0 gives x = 7.5

⇒ The price of a one-way ticket that will maximize revenue is $7.50

7 0

2 years ago

A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was co

Firlakuza [10]

Answer:

Step-by-step explanation:

1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

The correct option is D

2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

p = 70/100 = 0.7

q = 1 - p = 1 - 0.7 = 0.3

n = number of selected users = 50

Mean, u = np = 50×0.7 = 35

Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

x = number of internet users

The formula for normal distribution is expressed as

z = (x - u)/s

We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

P(x lesser than or equal to 25)

The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

Looking at the normal distribution table, the corresponding z score is 0.001

P(x lesser than or equal to 25) = 0.001

7 0

2 years ago