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1 year ago

Chris invested some money at a rate of 5% per year compound interest.

Mathematics

1 answer:

max2010maxim [7]1 year ago

6 0

Step-by-step explanation:

the answer is in the picture ☝️

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Two movie tickets and 3 snacks are $24. Three movies tickets and 4 snacks are $35. How much is a movie ticket and how much is a

Vsevolod [243]

Answer: A movie ticket is $9 while a snack is $2

Step-by-step explanation: We shall let a movie ticket be m while a snack is s. So, from the clues given, if two movie tickets and three snacks cost $24, we can write it as the following expression;

2m + 3s = 24

Also if three movie tickets and four snacks cost $35, we can as well write another expression as follows;

3m + 4s = 35.

Now we have a pair of simultaneous equations which are

2m + 3s = 24 ----------(1)

3m + 4s = 35 ----------(2)

We shall solve this by using the elimination method, since none of the variables has a coefficient of 1. We'll start by multiplying equation (1) by 3 and multiplying equation (2) by 2 (so as to eliminate the m variable)

2m + 3s = 24 -------- x3

3m + 4s = 35 ---------x2

We now arrive at the following

6m + 9s = 72--------(3)

6m + 8s = 70--------(4)

Subtract equation (4) from equation (3) and we arrive at

s = 2

Having determined that s equals 2 we can now substitute for the value of a into equation (1)

2m + 3s = 24

2m + 3(2) = 24

2m + 6 = 24

Subtract 6 from both sides of the equation

2m + 6 - 6 = 24 - 6

2m = 18

Divide both sides of the equation by 2

m= 9

Therefore one movie ticket costs $9 while one snack costs $2

6 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:

Learn more about inverse of the functionhttps://brainly.com/question/1632445.
Learn more about equation of circle brainly.com/question/1506955.
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

1 year ago

Read 2 more answers

the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard dev

Naya [18.7K]

Answer:

a) 2.84% probability that he is late for his first lecture.

b) 5.112 days

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 16, \sigma = 2.1$

a. Find the probability that he is late for his first lecture.

This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 16}{2.1}$

$Z = 1.905$

$Z = 1.905$ has a pvalue of 0.9716

1 - 0.9716 = 0.0284

2.84% probability that he is late for his first lecture.

b. Find the number of days per year he is likely to be late for his first lecture.

Each day, 2.84% probability that he is late for his first lecture.

Out of 180

0.0284*180 = 5.112 days

4 0

1 year ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago

Kate deposits $800 in a bank

Rina8888 [55]

Answer:

Step-by-step explanation:

Use formula

$I=P\cdot r\cdot t,$

where

I = interest

P = rpincipal

r = rate (as decimal)

t = time (in years)

In your case,

P = $800

r = 0.04 (4%)

t = 5,

$I=\$800\cdot 0.04\cdot 5\\ \\I=\$160$

6 0

1 year ago