All categories

1 year ago

Evaluate.

Mathematics

1 answer:

aleksandrvk [35]1 year ago

3 0

Answer: -2 4/15

Step-by-step explanation: To properly subtract, we need to find a common denominator. We can list the multiples of 3 and 5 to find the common denominator:

3: 3, 6, 9, 12, 15

5: 5, 10, 15

The first common multiple we see is 15.

Calculation:

3 · 3 = 9

9 + 1 = 10

3 1/3 = 10/3

5 x 5 = 25

25 + 3 = 28

-5 3/5 = -28/5

Now to convert into fifteen as the denominator:

-28/5 = -28 x 3/5 x 3 = -84/15

10/3 = 10 x 5/3 x 5 = 50/15

Now to subtract accordingly:

50/15 - 84/15 = -34/15

Final answer: -34/15 (can be reduced to -2 4/15).

You might be interested in

The value of a baseball player's rookie card began to increase once the player retired. When he retired in 1998 his card was wo

Sergio039 [100]

Y=2.05(X)+8.59
Yes because the value of y (value of card) changes as (x) amount of years pass by.
In 2009, the card is worth $31.14

6 0

2 years ago

Please help and EXPLAIN

creativ13 [48]

Okay, all of these pairs add up to 1 or 100%, except for one pair. I'll convert them all to decimals, so all must add up to 1.

3/8 = 0.375
0.625 + 0.375 = 1

62% = 0.62
0.38 + 0.62 = 1

7/8 = 0.875
0.875 + 0.125 = 1

70% = 0.70
1/3 <span>≈</span> 0.33
0.70 + 0.33 = 1.03

So the last pair of probabilities(70%, 1/3) does not belong with the other three because it does not add up to 1 like the others.

8 0

2 years ago

An automobile manufacturer claims that its car has a 35.935.9 miles/gallon (MPG) rating. An independent testing firm has been co

Serjik [45]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean.

For the given claim , we have

Null hypothesis : $H_0: \mu=35.9$

Alternative hypothesis : $H_a: \mu\neq35.9$

Since alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n=220 ;

Sample mean: $\overline{x}=35.6$ ;

Standard deviation: $s=2.2$

Test statistic for population mean:

$z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$z=\dfrac{35.6-35.9}{\dfrac{2.2}{\sqrt{220}}}\approx-2.02$

By using the standard normal distribution table of z , we have

P-value ( two tailed test ) : $2P(Z>|z|)=2(1-P(Z$

$=2(1-P(z$

Since , the P-value is greater than the significance level of $\alpha=0.02$ , it means we do not have sufficient evidence to reject the null hypothesis.