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1 year ago

5

Melissa buys 212 pounds of salmon and 114 pounds of swordfish. She pays a total of $31.25, and the swordfish costs $0.20 per pou

nd less than the salmon. What would be the combined cost of 1 pound of salmon and 1 pound of swordfish?

1 answer:

charle [14.2K]1 year ago

3 0

Answer: b

Step-by-step explanation:

You might be interested in

John's commute to work is 20kmhr while Sheri's commute is 500mmin. Who has the fastest commute to work in mihrif 1.61km=1mi? A S

Andreyy89

Answer:

A) Sheri has the faster commute by 6.2 miles/hr.

Step-by-step explanation:

Given

John's commute to work $=20\ km/hr$

Sheri's commute to work $=500\ m/min$

$1.61\ km = 1\ mile$

John's commute to work in miles per hour = $\frac{20\ km}{1 hr}\times \frac{1\ mile}{1.61\ km}= 12.42\ miles/hr$

Sheri's commute to work in miles per hour = $\frac{500\ m}{1\ min}\times \frac{1\ km}{1000\ m}\times \frac{1\ mile}{1.61\ km}\times \frac{60\ min}{1\ hr}= 18.63\ miles/ hr$

We can see that Sheri has a faster commute.

Difference between the rates = $18.63\ miles/ hr-12.42\ miles/hr=6.21\ miles/hr\approx 6.2\ miles/hr$

∴ Sheri has the faster commute by 6.2 miles/hr.

3 0

2 years ago

The Bay Area Online Institute (BAOI) has set a guideline of 60 hours for the time it should take to complete an independent stud

yarga [219]

Answer:

At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 60 \ hr$

The sample size is $n = 16$

The sample mean is $\= x = 68 \ hr$

The standard deviation is $\sigma = 20 \ hr$

The null hypothesis is $H_o : \mu = 60$

The alternative $H_a : \mu > 60$

Here we would assume the level of significance of this test to be

$\alpha = 5\% = 0.05$

Next we will obtain the critical value of the level of significance from the normal distribution table, the value is $Z_{0.05} = 1.645$

Generally the test statistics is mathematically represented as

$t = \frac{ \= x - \mu}{ \frac{ \sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 68 - 60 }{ \frac{ 20 }{\sqrt{16} } }$

$t = 1.6$

Looking at the value of t and $Z_{\alpha }$ we see that $t< Z_{\alpha }$ hence we fail to reject the null hypothesis

This means that there no sufficient evidence to conclude that it takes more than 60 hours to complete the course

So

At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

6 0

1 year ago

The table shows data gathered by an environmental agency about the decreasing depth of a lake during the first few months of a d

EastWind [94]

(0,346)(2,344.8)
slope = (344.8 - 346) / (2 - 0) = -1.2 / 2 = -0.6

y = mx + b
slope(m) = -0.6
use either of ur points (0,346)...x = 0 and y = 346
now we sub and find b, the y int
346 = -0.6(0) + b
346 = b

so ur equation is : y = -0.6x + 346

after 4 weeks....x = 4
y = -0.6(4) + 346
y = -2.4 + 346
y = 343.6 <=== after 4 weeks it will be 343.6

5 0

1 year ago

Which could be the graph of f(x) = |x - h| + k if h and k are both positive? On a coordinate plane, an absolute value graph has

Andrew [12]

Answer:

Step-by-step explanation:

f(x) = |x - h| + k has a vertex at (h, k), where both h and k are positive. Only

"On a coordinate plane, an absolute value graph has a vertex at (2, 1)" satisfies those requirements.

4 0

1 year ago

A teacher gives a test to a large group of students. The results are closely approximated by a normal curve. The mean is 74, wi

Alinara [238K]

Answer:

Step-by-step explanation:

Given that a teacher gives a test to a large group of students. The results are closely approximated by a normal curve

mu =74 and sigma =8

A grade starts from 100-8 = 92nd percentile

Z score for 92nd percentile = 1.405

X score = 74+8(1.405) = 85.24

--------------------

B cut off is to next 16%

Hence C would start for scores below 100-(8+16) = 76%

76th percentile = 0.705*8+74 =79.64

6 0

1 year ago