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Kitty [74]
3 years ago
14

The daily recommended allowance of vitamin C for six graders is 45 MG 1 orange has an about 75% of the recommended daily allowan

ce of five and see how many milligrams are in 150 guess I consider using the double number line

Mathematics
1 answer:
slamgirl [31]3 years ago
8 0

Answer:

So there are 33.75 MG in an orange.

Step-by-step explanation:

I think your full question in the photo below,

Here is my answer:

1 orange has an about 75% of the recommended daily allowance: 0.75*45 = 33.75 MG.

So there are 33.75 MG in an orange.

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On a piece of paper graph y=-2x-4
Snezhnost [94]

Answer:

Use a graphing calc or desmos to help guide you.

Step-by-step explanation:

3 0
3 years ago
Assume that y varies inversely with x. If y= 4 when x = 4, find y when x = 2. <br> y=[ ? ]
ahrayia [7]

Answer:

4

Step-by-step explanation:

y inversely proportional to x

y=1k/x

4=k/4

k=4×4

k=16

to find y

y=16/4

y=4

4 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
Sarah buys a shirt on sale at Jersey's. The original price was $28, but the shirt is on sale for 30% off. If there is a 5% sale
Helga [31]

Answer:

$20.58

Step-by-step explanation:

28 x 0.30 = 8.4

28 - 8.4 = 19.6

19.6 x 0.05 = 0.98

19.6 + 0.98 = 20.58

4 0
2 years ago
Please help -2=3x/5+1
Shalnov [3]

Answer:

X = -5

Step-by-step explanation:

Subtract one from both sides

- 3 = 3x / 5

multiply by 5 to remove fraction

-15 = 3x

divide by 3

x = -5

6 0
3 years ago
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