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4 years ago

234 divided by 33 I really need this answer doing school work

Mathematics

2 answers:

skad [1K]4 years ago

8 0

Answer:

234÷33= 7.09090909 or just round to 7.1

Elodia [21]4 years ago

4 0

Answer:

7.0909090909 hope this helps!!

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What is the slope of the line in the graph

nikitadnepr [17]

Answer:

well you need to attech the graph

Step-by-step explanation:

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2 years ago

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A store is having a Winter Blowout sale, where all their winter clothes are on sale for 75% off. Michael spent a total of $89.92

gogolik [260]

Answer:

$89.92 \div 25 = 3.5968 \times 100 = 359.68$

therefore the answer is 359.68$

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3 years ago

Is 1/8 - 10(3/4-3/8x)+5/8x equivalent to -1/8(59-35x)

madreJ [45]

Answer:

Yes

Step-by-step explanation:

1/8-10(3/4-3/8x)+5/8x

1/8-(30/4-30/8x)+5/8x

1/8-30/4+30/8x+5/8x

Lcm is 8

1-60+30x+5x/8

-59+35x/8

Factorise

-1/8(59-35x)

So when comparing the two,they are observed to be the same

5 0

3 years ago

Kayleigh has $4500 in a savings account at the bank that earns 0.8% interest per year. How much

Sloan [31]

Answer:

Kayleigh will have a total of $4608.

Step-by-step explanation:

First, you use the formula, I=PRT (Interest=Principal, Rate, Time), then you distribute the numbers: (I=4500x0.8%x3) when you multiply them all, you get $108, then you lastly add 108 to 4500, and you get your final answer of $4608.

7 0

3 years ago

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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 $2P(T$ 0 (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