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rjkz [21]
3 years ago
15

What single percentage change is equivalent to a 11% increase followed by a 15% increase?​

Mathematics
1 answer:
AlekseyPX3 years ago
7 0

Answer:

It's a total increase by 27,65%

Step-by-step explanation:

1,11 × 1,15 = 1,2765

1,2765 = +27,65%

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Brielle and her family went to our diner. They got ice cream cone, a brownie, and a cookie. How much did they spend on dessert?
Svetlanka [38]

Answer:

$6.83

Step-by-step explanation:

A brownie costs $3.45, an ice cream costs $2.43, and a cookie costs $0.95. So $3.45 + $2.43 + $0.95 = $6.83.  If you need to know the change if they paid with $10.00 also  . . . That would be $10.00 - $6.83 = $3.17 change.

7 0
3 years ago
(6.253•10^-2)(7.112x10^3) in scientific notation
zlopas [31]

Answer:

4.4471336 * 10^2.

Step-by-step explanation:

(6.253•10^-2)(7.112x10^3)

= 6.253*7.112 * 10^-2*10^3

= 44.471336 * 10^1

= 4.4471336 * 10^2.

7 0
3 years ago
HELP PLEASE ASAP
Tpy6a [65]

Answer: √20

Step-by-step explanation:

4.47213595 (the 5 repeats)

4 0
3 years ago
Factor each completely. <br><br>10a² - 9a + 2 <br><br>​
Sophie [7]

Answer:

a= \frac{2}{5} or \frac{1}{2}

Step-by-step explanation:

Equate the equation to zero

10a^{2}-9a+2=0

10a^{2}-5a-4a+2=0

5a(2a-1)-2(2a-1)=0

Write out the factors

(5a-2)(2a-1)=0

Then equate each factor to zero

5a-2=0 OR 2a-1=0

5a=2  OR  2a=1

a= \frac{2}{5}   OR a= \frac{1}{2}

a=\frac{2}{5} or \frac{1}{2}

3 0
3 years ago
Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.5 and a mean diameter of 205
Crank

Answer:

P(205-0.3=204.7

z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778

z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778

So we can find this probability:

P(-1.778

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

P = 1-0.9243 = 0.0757

Step-by-step explanation:

Let X the random variable that represent the diamters of interest for this case, and for this case we know the following info

Where \mu=205 and \sigma=1.5

We can begin finding this probability this probability

P(205-0.3=204.7

For this case they select a sample of n=79>30, so then we have enough evidence to use the central limit theorem and the distirbution for the sample mean can be approximated with:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}

And we can find the z scores for each limit and we got:

z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778

z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778

So we can find this probability:

P(-1.778

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

P = 1-0.9243 = 0.0757

6 0
3 years ago
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