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3 years ago

Can someone help please ?

Mathematics

1 answer:

MAVERICK [17]3 years ago

7 0

Answer:

B because Comp A is $60 for sure plus $42.95 per month and Comp B is $25 for sure and $49.95 per month

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If (6^2)^x equals 1, what is the value of x

ValentinkaMS [17]

Answer:

x = 0

Step-by-step explanation:

If p is a number other than zero, then

$p^x=1$

only if x=0.

8 0

3 years ago

A $3,500 loan, taken now, with a simple interest rate of 5% per year, will require a total repayment of $3,850. At what time t w

Alecsey [184]

Answer:

Roughly 2 years

Step-by-step explanation:

3500(1.05)^t=3850

(1.05)^t=1.1

t=1.953

So roughly 2 years

5 0

3 years ago

Read 2 more answers

Which system of equations could be graphed to solve the equation below? log Subscript 0. 5 Baseline x = log Subscript 3 Baseline

Vanyuwa [196]

You can use the fact that two expressions in equality can be considered to be equal to a third variable(not used in given context).

The system of equations that could be graphed to solve the equation given is

$y = \log_{0.5}(x)\\\\$
$y = \log_3(2+x)$

<h3>How can we form a system of equations from an equation?</h3>

Suppose the equation be $a = b\\$

Let there is a symbol $c$ such that we have $a = b = c$

It is because a and b are same measure (that is exactly what a = b means)

and we gave another name c to that measure.

Thus, we have

$a = c\\b = c$

in addition to $a = b\\$

<h3>Using the above method to find the system of equations needed</h3>

Since the given equation is $log_{0.5}(x) = log_2(2+x)$

The 2d graphs are usually expressed as $y = f(x)$ on X-Y plane.

Taking the equation's expressions equal to y, we get

$log_{0.5}(x) = log_2(2+x) = y$

or, we get system of equations as

$y = log_{0.5}(x)\\\\y = log_3{(2 + x)}$

Their graph is plotted below. The intersection point of both curves is the solution to the given equation as it satisfies both the equations of the system of equations formed from the given equation.

Thus,

The system of equations that could be graphed to solve the equation given is

$y = \log_{0.5}(x)\\\\$
$y = \log_3(2+x)$

Learn more about solutions to system of equations here:

brainly.com/question/14550337

4 0

3 years ago

Find a formula for the linear function with slope -9 and x-intercept 2.

gregori [183]

Hello :
<span> formula for the linear function with slope -9 and x-intercept 2 is :
y - 0 = -9(x-2)
y= -9x +18

</span>

4 0

4 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, <u><em>the test statistics</em></u> = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

3 years ago