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taurus [48]
4 years ago
10

Help me, cause I'm tuck

Mathematics
1 answer:
r-ruslan [8.4K]4 years ago
8 0

Answer:

The answer would be 1.33 x 10^{4}

Step-by-step

1.28 x 10^{4} = 12800

5 x 10^{2} = 500

12800 + 500 = 13300

13300 = 1.33 x 10^{4} in scientific notation

Hope this helps:)

Sorry, I got the wrong answer earlier but I fixed it:)

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Giselle works as a carpenter and as a blacksmith. She earns $ 2 0 $20dollar sign, 20 per hour as a carpenter and $ 2 5 $25dollar
dem82 [27]
She worked as a carpenter for 12 hours and as a blacksmith for 18 hours.

Assuming you mean she earned $20 as a carpenter and $25 as a blacksmith per hour, with a total of 30 hours for $690,
let c represent carpenter hours and b for blacksmith hours.

20c + 25b = 690
c + b = 30
Subtract b from each side so that c = 30 - b
Plug this value into the first equation

20(30 - b) + 25b = 690
600 - 20b + 25b = 690
600 + 5b = 690
5b = 90
b = 18

To find c, plug this value of b into the other equation
c + 18 = 30
c = 12


3 0
3 years ago
Read 2 more answers
Can yall help me out​
Montano1993 [528]

Answer:

C

Step-by-step explanation:

The range is the values of y that extend from the vertex downwards

The vertex = (- 2, - 1) → y = - 1

range is (- ∞, - 1 ]

7 0
3 years ago
A researcher wishes to see if the average number of sick days a worker takes per year is less than 5. A random sample of 30 work
Lana71 [14]

Answer:

No. At a significance level of 0.01, there is not enough evidence to support the claim that the average number of sick days a worker takes per year is significantly less than 5.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average number of sick days a worker takes per year is significantly less than 5.

Then, the null and alternative hypothesis are:

H_0: \mu=5\\\\H_a:\mu< 5

The significance level is 0.01.

The sample has a size n=30.

The sample mean is M=4.8.

The standard deviation of the population is known and has a value of σ=1.2.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.2}{\sqrt{30}}=0.219

Then, we can calculate the z-statistic as:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{4.8-5}{0.219}=\dfrac{-0.2}{0.219}=-0.913

This test is a left-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z

As the P-value (0.181) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the average number of sick days a worker takes per year is significantly less than 5.

3 0
3 years ago
The float committee made a scale model of their float before they started construction. The float committee used the scale 1 inc
igor_vitrenko [27]

Answer:

6

Step-by-step explanation:

7 0
3 years ago
ASAP ANSWER PLEASE :)
Hunter-Best [27]

It is 3.33%

Origninally, the decimal continues with repetitive 3s.

4 0
3 years ago
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