Answer:
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
The proportion of infants with birth weights between 125 oz and 140 oz is
This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So
X = 140
has a pvalue of 0.9772
X = 125
has a pvalue of 0.8413
0.9772 - 0.8413 = 0.1359
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
Answer:
$4.88
Step-by-step explanation:
Jessica made $16 mowing a neighbors yard
She brought a craft kit for $7.37
16-7.37
= 8.63
She used $3.75 to buy food for a food drive
8.63-3.75
= 4.88
Hence the amount Jessica put in the savings account is $4.88
Answer: 14x^2-93xy+60y^2 Hope that helps!
Step-by-step explanation:
1. Expand by distributing terms
(20x-12y)(x-4y)-(3x-4y)(2x+3y)
2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)
3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)
4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2
5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)
6. Simplify.
And your answer would be 14x^2-93xy+60y^2
The answer would be 200,000.
This is because you go to the hundred thousands' place and look one place to the right. Since there is a 6 and 6>5 you round up to 2 making it 200,000. Hope this helps!
6z - 4 = 60
6z = 64
z = 10.66666666666666 (repeating)
