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1 year ago

The relationship between goby fish and striped shrimp are truly/A symbiotic, for neither/B can survive/C

1 answer:

7 0

The relationship between goby fish and striped shrimp are truly/A symbiotic, for neither/B can survive/C without/D the other. No error/E

The correct option is A.

Yes, it should be is.
The sentence is referring to the relationship between the fish, not the fish themselves.
Relationship is singular.

What is symbiotic explain?

Symbiosis, any of several living arrangements between members of two different species, including mutualism, commensalism, and parasitism.
Both positive (beneficial) and negative (unfavourable to harmful) associations are therefore included, and the members are called symbionts.

Learn more about symbiotic

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the walters backyard pool is rectangular the pool measures 10 feet long and 8 feet wide. If the length of the backyard is 35 fee

Vlad1618 [11]

Answer:28

Step-by-step explanation:

10×

5 0

3 years ago

Read 2 more answers

If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

Ilya [14]

This is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the "common difference" "d".

now, we know the 17th term is 34, let's go to the 15term then

34 + d
34 + d + d
34 + d + d + d
34 + d + d + d + d
34 + d + d + d + d + d
34 + d + d + d + d + d + d
34 + d + d + d + d + d + d + d
34 + d + d + d + d + d + d + d + d

now, notice, we got to the 15th term, which is 34 + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

$\bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}$

ok, now that we know what the common difference is, let's find the first term in the sequence using the 7th term of 34,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=5\\
a_7=34
\end{cases}
\\\\\\
a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5
\\\\\\
34=a_1+30\implies \boxed{4=a_1}$

and now let's use those 2 found folks, to get the 21st term,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=4\\
d=5\\
n=21
\end{cases}
\\\\\\
a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104$

3 0

3 years ago

New york sales tax is charged at the rate of 6%. find the tax and the total price you would pay for an $860 stereo.

Serhud [2]

To calculate the tax multiply $860 by .06 (6%) to get $34.40.
To calculate the total amount you'll pay add $34.40 to $860 to get $894.40.

8 0

3 years ago

Yasmin purchased 6 heads of cabbage that each weighed 2 3/8 pounds. How much did the cabbage weigh all together?

nika2105 [10]

Answer:

14 $\frac{1}{4}$ pounds

Step-by-step explanation:

Here we have to use multiplication.

Yamin purchased 6 heads of cabbage.

The weight of each cabbage = 2 $\frac{3}{8}$

First let's convert the mixed fraction 2 $\frac{3}{8}$ to improper fraction

2 $\frac{3}{8}$ = $\frac{19}{8}$

The weight of 6 heads of cabbage = 6 x $\frac{19}{8}$

= 114/8

= 14 $\frac{2}{8}$

We can reduce 2/8 = 1/4

= 14 $\frac{1}{4}$

Thank you.

3 0

3 years ago

The General Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours

DerKrebs [107]

Answer:

Yes, we can conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

Step-by-step explanation:

H0 : σ² = 2.66²

H1 : σ² < 2.66²

X²c = (n - 1)*s² ÷ σ²

sample size, n = 40

Sample standard deviation, s = 1.9

X²c = ((40 - 1) * 1.9²) ÷ 2.66²

X²c = 140.79 ÷ 7.0756

X²c = 19.897

Using a confidence level of 95%

Degree of freedom, df = n - 1 ; df = 40 - 1 = 39

The critical value using the chi distribution table is 25.6954

Comparing the test statistic with the critical value :

19.897 < 25.6954

Test statistic < Critical value ; Reject the Null

Hence, we can conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

8 0

3 years ago