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jenyasd209 [6]
3 years ago
7

3 is to 14 as 12 is to what number? O 3.5 O 2.6 O 42 O 56

Mathematics
2 answers:
snow_tiger [21]3 years ago
6 0
12 is to 56 is the correct answer or just option D lol :)
Tamiku [17]3 years ago
6 0

Answer:

12 is to 56

Step-by-step explanation:

hope it helped have a good day :))

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How do I factor out the coefficient of the variable ?<br><br> 3/10y-2/5
Sindrei [870]
Add 3/10 to 2/5 to factor it out
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3 years ago
Alex has a block of wood that is in the shape of a prism with the dimensions shown. He cuts a 10-cm square hole through the cent
eimsori [14]

Answer:

Step-by-step explanation:

I think your question is missing key information and the one below is similar to yours. Hope it will find you well:

"Alex has a block of wood that is in the shape of a rectangular prism dimensions 70 cm by 50 cm by 30 cm.  He cuts a 10-cm square hole through the center of the prism. What is the volume of the remaining solid?

Here is my answer:

The volume of the rectangular prism: 70∗50∗30=105000 cm^{3}

The volume  in the center (intersection of “square holes”): 10∗10∗10=1000 cm^{3}

=> the volume of the remaining solid = The volume of the rectangular prism - The volume  in the center = 105000 - 1000 = 104000cm^{3}

8 0
3 years ago
Read 2 more answers
If it starts out at -10 degrees, what will the temperature be if it rises 30 degrees?
ladessa [460]

-10 + 30

so the answer is 20 degrees

6 0
3 years ago
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
5. Suppose that 15% of a 12-gal solution is bleach and
puteri [66]

Step-by-step explanation:

15 percent of solution = 12 × 0.15 = 1.8 Gallons of bleach

12 - 1.8 = 10.2 Gallons of water

5 0
3 years ago
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