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Marysya12 [62]
3 years ago
6

The price of a car increased by 7% which expression can be used to calculate the new price of the car

Mathematics
1 answer:
julsineya [31]3 years ago
6 0

Answer: P * ( 1 + 7%)

Step-by-step explanation:

You included no options but the expression should go something like this:

Pn = P * ( 1 + 7%)

Where Pn is the new price

P is the current price

<em>The above formula will show the new price given the current price. </em>

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Answer to 42.7 -(-12.4)
Aleks [24]
<span>42.7−<span>(<span>−12.4</span>)

</span></span><span>=<span>42.7−<span>(<span>−12.4</span>)

</span></span></span><span>=<span>42.7+12.4

</span></span><span>=<span>55.1</span></span>
4 0
3 years ago
Which of the following is the radical expression of a to the four ninths power ?
LuckyWell [14K]

Given expression in exponential form : a^{\frac{4}{9} }.

We need to convert it into radical form.

<em>Please note: When we convert an exponential to radical form, the top number goes in the exponent of the term and bottom number of the fraction goes in the radical sign to make it nth radical.</em>

We can apply following rule:

(a)^{\frac{m}{n}}= \sqrt[n]{x^m}.

Therefore,

a^{\frac{4}{9} }= \sqrt[9]{a^4}.

Therefore, correct option is : D. ninth root of a to the fourth power.

6 0
3 years ago
A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life
alisha [4.7K]

Answer:

a)

The null hypothesis is H_0: \mu \leq 10

The alternative hypothesis is H_1: \mu > 10

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

H_0: \mu \leq 10

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

H_1: \mu > 10

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that \mu = 10

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that n = 45, X = 10.5, s = 1.8

Then

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}

t = 1.86

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

5 0
2 years ago
Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).
fomenos

Answer:

\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is 2a=|4--10|.

\implies 2a=|14|

\implies 2a=14

\implies a=7

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is (\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)

We need to use the relation a^2+b^2=c^2 to find b^2.

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

c=|6--3|=9

\implies 7^2+b^2=9^2

\implies b^2=9^2-7^2

\implies b^2=81-49

\implies b^2=32

We substitute these values into the standard equation of the hyperbola to obtain:

\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1

\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1

7 0
3 years ago
Write as an algebraic expression: <br><br> 1) a% of 80 <br><br> 2) 17% of b <br><br> 3) a% of b
Anna35 [415]
1) 4a/5
a% of 80 = a% (80) = (a/100)(80) = 80a/100 = 4a/5

2) 17b/100 or 0.17b
17% of b = (17/100)(b) = 17b/100 [ = 0.17b ]

3) ab/100
a% of b = (a/100)(b) = ab/100
5 0
3 years ago
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