The amount of money in Malik's bank account 3 years ago was $1,000. This year his balance is $1,900.

When an object falls freely, the formula for the distance it falls is d= 4.9t^2 where d is

german

Answer:

a) 122.5 m ; b) 10 s

Step-by-step explanation:

Just use the given equation and plug in what you know.

a)

d = 4.9t^2

d = 4.9(5^2)

d = 4.9 * 25

d = 122.5 m

b)

d = 4.9t^2

490 = 4.9t^2

100 = t^2

t = 10 s

3 0

3 years ago

Read 2 more answers

Evaluate <br><br> 2r(t + 7) if r = 4 and t = 2.

Serhud [2]

The answer is 72 all u need to do plug in the numbers where the letters are

6 0

3 years ago

The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in

Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 64 inches

$\sigma$ = standard deviation = 4 inches

n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{63-64}{\frac{4}{\sqrt{4} } }$ ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

= 1 - 0.69146 = 0.30854

Hence, it is 30.85% likely that average height would be shorter than 63 inches.

7 0

3 years ago

A real estate developer is considering investing in a shopping mall on the outskirts of Atlanta, Georgia. Three parcels of land

Arada [10]

Answer:

Step 1: The null and alternative hypothesis for the one way analysis of variance is,

H₀ : There is no difference in the means of each area.

H₁ : At least one mean is different.

The objective of this question is to test whether the income of the three areas are significantly different.

Step 2: The problem states to use 0.05 significance level.

Step 3: The test statistic follows the F-distribution.

Step 4: Decision rule is based on the critical value.

Numerator degrees of freedom is,

K - 1 = 3 - 1

= 2

Denominator degrees of freedom is 2.

n - k = 12 - 3

= 9

Using the F-distribution table for ∝=0.05 with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to reject H₀ if the computed value of F exceeds 4.26.

Step 5: Calculations.

Using Excel follow the steps mentioned below.

1. Import or type the data into spreadsheet.

2. Data--------> Data Analysis----------> Anova: Single Factor, click OK.

3. Select Input Range, make a tick mark on Labels in the first row, enter 0.05 for alfa.

4. Click OK.

The obtained output for the one-way ANOVA is,

[ find the figure in attachment]

From the obtained output, the computed test statistic value, F = 14.18 , which is greater than the critical value of 4.26. So, reject H₀

Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .

4 0

3 years ago

Help with Algebra 2: 1. <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B7%5D%7Bx%5E5%7D%20%7D%7B%5Csqrt%5B4%5D%7Bx%5E2%7D%

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

1. Given the expression

$\dfrac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }$

Note that

$\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}$

When dividing $\sqrt[7]{x^5}$ by $\sqrt[4]{x^2},$ we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so