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3 years ago

Suppose an investment of $8,200 doubles in value every 7 years. How much is the investment worth after 28 years?

Mathematics

2 answers:

lutik1710 [3]3 years ago

6 0

$229,600 is how much money you would have in 28 years.

Rudiy273 years ago

5 0

8200 * 2= 16400
16400 * 2= 32800
32800 * 2= 65600
65600 * 2 = 131200

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Jean is 25 years older than her daughter Charlotte. In ten years, Jean will be twice Charlotte's age. How old are Jean and Charl

allsm [11]

Charlotte is 15 while jean is 40

3 0

3 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

3 years ago

Which figure appears to show a line with a slope of 1?

gulaghasi [49]

Answer:

I believe it is figure 1

7 0

3 years ago

Read 2 more answers

For the following hypothesis test, determine the null and alternative hypotheses. Also, classify the hypothesis test as two-tail

bagirrra123 [75]

Answer:

Option B:

$H_0: \mu = 22.1$

$H_a: \mu \neq 22.1$

Classification:

The hypothesis test is Two-tailed.

Step-by-step explanation:

The mean length of imprisonment for motor-vehicle theft offenders in this country is 22.1 months.

This means that the null hypothesis is that the mean is of 22.1 months, that is:

$H_0: \mu = 22.1$

A hypothesis test is to be performed to determine whether the mean length of imprisonment for motor-vehicle theft offenders in this city differs from the national mean of 22.1 months.

At the alternate hypothesis, we test if this mean is different of 22.1, that is:

$H_a: \mu \neq 22.1$

Which means that the answer is given by option b).

Which of the following is the correct classification of the hypothesis test?

We test if the mean is different from a value, which means that the hypothesis test is Two-tailed.

7 0

3 years ago

0.336 divided by 0.2

Lilit [14]

1.68 I hope it helps!

7 0

3 years ago