Can you simplify (rt(2)+rt(6))/2?

Mariko and her friend spent $24.50 on

jolli1 [7]

For the lunch it would be 20$

4 0

3 years ago

Read 2 more answers

If Julie invests $9,250 at a rate of 7%, compounded weekly, find the value of the investment after 5 years.

Reil [10]

Answer:

Option D is correct.

Step-by-step explanation

Principal = $9250

rate of interest = 7% or 0.07

time = 5 years or 260 weeks

[ Since there are 52 weeks in a year . for 5 years it will be 5x52=260 weeks]

Applying the formula

Amount after t years = $P(1+\frac{r}{n} )^{nt}$

where P = principal

r = rate % in decimals

n= number of times in a year

t = times ( in years)

plugging the values in the formula

Amount = $9250(1+\frac{0.07}{52} )^{(52X(5)}$

= $9250(1+0.001346 )^{(260)}$

= $9250(1.001346 )^{(260)}$

= 9250(1.418733588)

=$13123.29

5 0

3 years ago

Simplify the expression below. (5x)2

Anarel [89]

10x ..................

6 0

3 years ago

A Food Marketing Institute found that 29% of households spend more than $125 a week on groceries. Assume the population proporti

ella [17]

Using the normal distribution, there is a 0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

The estimate and the sample size are:

p = 0.29, n = 207.

Hence the mean and the standard error are given as follows:

$\mu = p = 0.29$ .

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.29(0.71)}{207}} = 0.0315$ .

The probability that the sample proportion of households spending more than $125 a week is less than 0.31 is the <u>p-value of Z when X = 0.31</u>, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.31 - 0.29}{0.0315}$

Z = 0.63

Z = 0.63 has a p-value of 0.7357.

0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

6 0

2 years ago

Xander needs to collect at least 120 cans for a food drive to earn community service credit. He has already collected 64 items.

Aleksandr [31]

Answer:

c ≥ 56 is the REQUIRED INEQUALITY.

Step-by-step explanation:

Here, the given question is INCOMPLETE.

Xander needs to collect at least 120 cans for a food drive to earn community service credit. He has already collected 64 items. Choose the inequality and solution to represent the number of cans, c, that Xander must still collect.

Now, here:

The number of cans Xander needed to collect = At least 120

The number of items already collected = 64

c: the number of cans, c, that Xander must still collect.

Now, the number of cans to be collected - Cans already collected

= 120 - 64 = 56

So, the number of can he must collect to make a TOTAL OF AT LEAST 120 cans = 56 cans

⇒ The number of cans to be collected ≥ 56 cans

⇒c ≥ 56 cans

or, c ≥ 56 is the REQUIRED INEQUALITY.

3 0

3 years ago