Answer:
Following are the responses to the given question:
Step-by-step explanation:

Answer:
ΔABC ~ ΔDEC
Step-by-step explanation:
Given : DE║AB
Statements Reasons
1). DE║AB 1). Given
2). ∠CDE ≅ ∠CAB 2). Corresponding angles
[Since DE║AB and AC is the transverse]
3). ∠CED ≅ ∠CBA 3). Corresponding angles
[Since DE║AB and BC is the transverse]
4). ΔABC ~ ΔDEC 4). By AA property of similarity
Hence ΔABC is similar to ΔDEC.
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%.
Jimmy asked me for 8 of my 17 cookies so i gave them to him which left me with nine so I went to the store and got 5 which gives me 14
I really dont get what your asking but I hope this helps you. If it doesn't please ask me for help ;)
Answer:
0.1587
Step-by-step explanation:
Here, mean=μ=100 and standard deviation=σ=16.
We have to find P(average MDI scores of 64 children > 102)=P(xbar>102).
n=64
μxbar=μ=100
σxbar=σ/√n=16/√64=16/8=2
P(xbar>102)=P((xbar-μxbar)/σxbar>(102-100)/2)
P(xbar>102)=P(z>1)
P(xbar>102)=P(0<z<∞)-P(0<z<1)
P(xbar>102)=0.5-0.3413
P(xbar>102)=0.1587
Thus, the probability that the average is greater than 102 is 15.87%