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Inessa [10]
2 years ago
15

7x10 to the power of 5​

Mathematics
1 answer:
Dafna11 [192]2 years ago
6 0

Answer:

700,000

Step-by-step explanation:

The exponent is 5, making it 10 to the power of 5. As the exponent is positive, the solution is a number greater than the origin or base number. To find our answer, we move the decimal to the right 5 times.

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Como soluciono equaciones =0​
Rom4ik [11]

Step-by-step explanation:

ما زلت لا أستطيع أن أخبرني بما تحاول أن تخبرني به ولد أو بنت آسف على تحمله

8 0
3 years ago
N and (2 + 6)<br> What is the answer
AleksandrR [38]

Answer:

8n

Step-by-step explanation:

distribute n between 2 and 6 then add

3 0
3 years ago
A rectangle has a height of 4w^3 and a width of 5w^2-3w-4.
Thepotemich [5.8K]

Answer:

Step-by-step explanation:

Area of the entire rectangle = height * width

= 4w³ *( 5w²-3w-4)

= 4w³ * 5w²- 4w³ *3w - 4w³ *4

=20w⁵ - 12w⁴ - 16w³

6 0
3 years ago
Read 2 more answers
Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s
stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have n_{1} = 25, \bar{x}_{1} = 2.04, \sigma_{1} = 0.010 and n_{2} = 20, \bar{x}_{2} = 2.07, \sigma_{2} = 0.015. The pooled estimate is given by  

\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative).  

The test statistic is T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}} and the observed value is t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257. T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value t_{0} falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 2P(T0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}, i.e.,

-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}

where t_{0.025} is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

8 0
3 years ago
Pls help me with my math i wanna play a p e x
Brrunno [24]

Answer:

THE FIRST ONE 0.018 IS 18/1000

Step-by-step explanation:

Q 2 is 4 13/20

3 0
3 years ago
Read 2 more answers
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