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3 years ago

15

A certain species of fish costs $2.99 each. You can spend at most $35. How many of this type of fish can you buy for your aquari

um?

1 answer:

Burka [1]3 years ago

8 0

Answer:

11

Step-by-step explanation:

2.99 x 11 = 33.89

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Only 18 people are allowed on a ride at the same time. There are 157 people waiting in line. How many groups of riders will ther

IrinaK [193]

Answer:

Step-by-step explanation:

Divide 157 by 8 to see how many groups there are.

157/18=8r13. So you can round up to 9 groups of riders as there is a max of 18 at a time.

7 0

3 years ago

Number 9 please. I need help

Sloan [31]

(1/12)X = 6800

6800/ (1/12) = X

X = 81600 (hours)

6800 (6.8 * 10^3) divided by 1/12 (12^-1) equals 81600. It will take 81600 hours for the pool to completely empty at a rate of 0.833333333 or 12^-1 gallons per second.

8 0

3 years ago

Twice the difference of a number and 1 is 4 more than that number. find a number using polyas 4 steps

professor190 [17]

Answer:

x=3

Step-by-step explanation:

2(x-1)=4

2x-2=4

2x=4+2

2x=6

x=6/2=3

8 0

3 years ago

Find the probability of having 2,3, or 4 successes in five trials of a binomial experiment in which the probability of success i

Georgia [21]

Answer:

65.3%

Step-by-step explanation:

Binomial distribution PMF below (along with a description)

the probabilty of getting 2, 3, or 4 successes in 5 trials is just

p(2)+p(3)+p(4)

p(2)= 5C2*.4²*(1-.4)³ = .3456

p(3)= 5C3*.4³*(1-.4)²= .2304

p(4)= 5C4*.4⁴*(1-.4)=.0768

.3456+.2304+.0768= .6528

6 0

3 years ago

A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand

sesenic [268]

Answer:

The 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =80$ represent the sample mean 1

$\bar X_2 =75$ represent the sample mean 2

n1=25 represent the sample 1 size

n2=36 represent the sample 2 size

$\sigma_1 =5$ sample standard deviation for sample 1

$\sigma_2 =3$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =80-75=5$

Since the Confidence is 0.94 or 94%, the value of $\alpha=0.06$ and $\alpha/2 =0.03$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that $z_{\alpha/2}=1.88$

The standard error is given by the following formula:

$SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118$

Confidence interval

Now we have everything in order to replace into formula (1):

$5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898$

$5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102$

So on this case the 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

7 0

3 years ago