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amid [387]
2 years ago
6

What expression has the estimated product of 60

Mathematics
1 answer:
balu736 [363]2 years ago
6 0

Answer:

30x2

Step-by-step explanation:

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Simplify the following expression. (1/7 + 3/8) + (2/9x - 1/8)
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2/9x+11/28

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A loan of $8 000 was repaid in 2 years in
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P = 8000

t = 2

I = p × r × t

400 = 8000 × r/100 × 2

r = 400/160 per year

r = 40/16 = 2.5 %

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The $200 digital camera Andres purchased was on sale for 15% off. How much was the discount amount?
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30

Step-by-step explanation:

200*0.15=30

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N.<br> Solve the expression.<br> Use PEMDAS<br> [2.1+(9.2 x 3.3)] x 0.8=<br> -
densk [106]

Answer:

  25.968

Step-by-step explanation:

The Google calculator will reliably use PEMDAS to evaluate an expression.

 = (2.1 +30.36)×0.8

 = 32.46×0.8

 = 25.968

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6 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

8 0
3 years ago
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