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Rina8888 [55]
3 years ago
5

SOMEONE HELP PLEASEEE I NEED ANSWERS "To express each product as a numeral, how many places to the right must you move the decim

al point? Express each product as a numeral." a) 5.9x10^8 b)6.35x10^4. C) 9.01x10^6 D) 1.12x10^3 e) 8.3x10^5 f) 4.592x10^11
Mathematics
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

590000000

63500

9010000

1120

830000

459200000000

Step-by-step explanation:

Express each product as a numeral."

a) 5.9x10^8 = 5.9 * 100000000 = 590000000

b)6.35x10^4 = 6.35 * 10000 = 63500

C) 9.01x10^6 ) 9.01 * 1000000 = 9010000

D) 1.12x10^3 = 1.12 * 1000 = 1120

e) 8.3x10^5 = 8.3 * 100000 = 830000

f) 4.592x10^11). 4.592 * 100000000000 = 459200000000

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ExtremeBDS [4]
First you have to turn pounds into ounces and its 16 oz for 1 pound.

Now you need to multiply 16 by 5.
16×5=80

Now you add the 4 oz to the 80 oz.
80+4=84

Now you need to divide $26.88 by the 84 oz.
26.88÷84=0.32 

Then you finally get $0.32 per 1 oz.

I really hope that helped : )
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3 years ago
Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio
Kryger [21]

Answer:

From the given data, there is not enough evidence to prove that there is a statistically significant difference between the two population means

Step-by-step explanation:

The number of younger moms, in the study = 837

The average weight gain of younger moms, \overline x₁ = 30.67 pounds

The standard deviation of the weight gain of younger moms, s₁ = 14.69 pounds

(30.67 - 28.52)/√((14.69^2)/837 + 13²/143))

The number of younger moms, in the study = 143

The average weight gain of mature moms, \overline x₂ = 28.52 pounds

The standard deviation of the weight gain of mature moms, s₂ = 13 pounds

The test statistic for the difference in two populations is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_1^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

Therefore, we get;

t=\dfrac{(30.67-28.52)}{\sqrt{\dfrac{14.69^{2} }{837}-\dfrac{13^{2}}{143}}} \approx 1.7919

The test statistic ≈ (1.7919)

Using a graphing calculator, we get;

The critical-t = ±1.971379, p = 0.07459697

Therefore, given that the test statistic, (1.7919), < critical-t (0.07459697), we fail to reject the null hypothesis, therefor, the given data does not provide convincing evidence that there is a significant difference between the two population means

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