Answer:
De value of k is 9
Step-by-step explanation:
6,-1
b(1,3)
(k,8)
am using bodmas
a6-1×1=6a
b1×3=3b
abc=8
6a+3b+8=17-8
=9
I think the answer is 176
Answer:
6x^2+xy-35y^2-2x
Step-by-step explanation:
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Answer:
It is likely that the birth weight of a random baby boy will be between 3.2 and 3.4 kg because the probability of this event is large enough.
Step-by-step explanation:
Population mean=μ=3.3.
S.E=0.1.
n=36.
If the probability of the birth weight of a random baby boy will be between 3.2 and 3.4 kg is larger than the it will be likely. The probability can be calculated by normal distribution because sample size is large enough.
Z-score for 3.2 kg=3.2-3.3/0.1=-1
Z-score for 3.4 kg=3.4-3.3/0.1=1
P(-1<Z<1)=P(-1<Z<0)+P(0<Z<1)
P(-1<Z<1)=0.3413+0.3413
P(-1<Z<1)=0.6826
The probability of the birth weight of a random baby boy will be between 3.2 and 3.4 kg is 68.26%. So. it is likely that the birth weight of a random baby boy will be between 3.2 and 3.4 kg as the probability is large enough.
Answer:
4.9*1/1000
is 4.9/1000 which is
0.049!
Step-by-step explanation: