15.
4(x-2) = 2(x-4)+2x
i used the distributive property on the parantheses to then get...
4x-8 = 2x-8+2x
I combined the like terms to then get...
4x-8 = 4x-8
+8 +8
4x = 4x
which simplifies to....
x=x
so the answer to question 15 is x=x
17.
-4(p+1) = -2(8-2p)
I used the distributive property on the parantheses....
-4p-4 = -16+4p
+4 +4
-4p = -12+4p
-4p -4p
-8p = -12
÷-8 ÷-8
p = 1.5
so the answer to question 17 is p = 1.5
hope this helps!!!!
Answer:
(a) The data are not sufficient to demonstrate that the herb has a significant effect on memory.
(b) Cohen's d = 5.07
Step-by-step explanation:
(a) Null hypothesis: The herbal supplement (ginkgo biloba) can impact human memory
Alternate hypothesis: The herbal supplement (ginkgo biloba) cannot impact human memory.
Z = (sample mean - population mean) ÷ sd/√n
Sample mean = 76, population mean = 70, sd = 15, n = 25
Z = (76 - 70) ÷ 15/√25 = 6 ÷ 3 = 2
Using a 0.05 significance level for a two tailed test, the critical value is 1.96
For a two tailed test, the region of no rejection of the null hypothesis lies between -1.96 and 1.96
Conclusion: Since the test statistic (Z = 2) falls outside the region bounded by -1.96 and 1.96, reject the null hypothesis. The herbal supplement (ginkgo biloba) cannot impact human memory.
Hence, the data are not sufficient to demonstrate that the herb has a significant effect on memory.
(b) Cohen's d = M/sd = 76/15 = 5.07
Answer:
27 is 45% of 60
Step-by-step explanation:
thank u hope it helps you
3. The original sequence
TAC - CGC - TTA - CGT - CTG - ATC - GCT
codes for
tyr - arg - leu - arg - leu - ile - ala
while the mutated sequence codes for
TAC - CGC - TTA - TTA - TTA - CGT - G<u>CT</u> - <u>G</u>CT - ATC - GCT
tyr - arg - leu - leu - leu - arg - <u>ala</u> - ala - ile - ala
There are several frameshift mutations involved here:
• the first inserts 6 bases (TTA - TTA)
• the second inserts 1 base (G) before the CTG triplet (underlined)
• the third inserts 2 bases (CT) after the CTG triplet
4. The original sequence is the same as before. The mutated sequence
TAC - CGC - TAA - TTA - TTA - CGT - G<u>CT</u> - <u>G</u>CT - ATC - GCT
codes for
tyr - arg - STOP - leu - leu - arg - ala - ala - ile - ala
Then
• there is a (nonsense) point mutation that swaps T for A in the original TTA triplet (nonsense since it produces a stop codon that would halt replication/expression)
• there is a frameshift mutation that inserts 3 bases (TTA)
as well as two other frameshift mutations that also occurred in the previous part.