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2 years ago

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Mike has a net spendable income of $1,400. He decides to set up a budget before looking for an apartment or a car. He sets up hi

s budget and finds that he has lots of money left over. He puts the extra money into entertainment.

2 answers:

forsale [732]2 years ago

5 0

I don't see a question here :/

nadya68 [22]2 years ago

5 0

Stupid kid you forgot to put the choices

You might be interested in

Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo

Zepler [3.9K]

Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period 2 is 9

Forecast can be given by

$F_{t+1}=F_t+\alpha (D_t-F_t)$

where

$F_{t+1}=Future Forecast$

$F_t=Present\ Period\ Forecast$

$D_t=Present\ Period\ Demand$

$\alpha =smoothing\ constant$

$F_{t+1}=5+0.2(7-5)$

$F_{t+1}=5.4$

Forecast for Period 3

$F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})$

$F_{t+2}=5.4+0.2\cdot (9-5.4)$

$F_{t+2}=6.12$

8 0

2 years ago

One nanometer equals about

Orlov [11]

Answer:

$1\ \text{nanometer}=4\times 10^8$

Step-by-step explanation:

Given : One nanometer equals about 0.00000004 inches.

To find : When writing this decimal as a single-digit integer multiplied by a power of ten, the single digit integer is and the power of ten ?

Solution :

$1\ \text{nanometer}=0.00000004$

$1\ \text{nanometer}=\frac{00000004}{100000000}$

$1\ \text{nanometer}=4\times 10^8$

Therefore, the decimal as a single-digit integer multiplied by a power of ten is $1\ \text{nanometer}=4\times 10^8$

4 0

2 years ago

Please answer these questions thanks

xz_007 [3.2K]

Well...
$6.50/64= about $0.10
assuming that the cost of the 48oz bottle is $4.20,
$4.20/48= about $0.08

the 48oz is better buy, as it is $0.08/oz
(/ means per)

7 0

2 years ago

A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year

uysha [10]

Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that $p = 0.8$ .

10 students, so $n = 10$

We have to find:

$P(X \geq 9) = P(X = 9) + P(X = 10)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684$

$P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758$

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that $p = 0.3758$

Two groups, so $n = 2$

We have to find P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691$

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

5 0

2 years ago

Read 2 more answers

The center of a hyperbola is (−2,4) , and one vertex is (−2,7) . The slope of one of the asymptotes is 1/2 .

maxonik [38]

Answer:

<h2>The equation is $\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1$ .</h2>

Step-by-step explanation:

The equation of a hyperbola is represented by $\frac{(y - k)^{2} }{b^{2} } - \frac{(x - h)^{2} }{a^{2} } = 1$ , where (h, k) is the center of the hyperbola.

As per the given condition, h = -2 and k = 4.

Thus, the equation becomes $\frac{(y - 4)^{2} }{b^{2} } - \frac{(x + 2)^{2} }{a^{2} } = 1$ .

One vertex is (-2, 7).

Hence, putting x = -2 and y = 7, we get $b^{2} = 9$ .

The slope of the asymptote is $\frac{b}{a}$ .

Hence, $\frac{b^{2} }{a^{2} } = \frac{1}{4} \\a^{2} = 36$ .

Thus, the equation is $\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1$ .

7 0

2 years ago