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3 years ago

A package being delivered in the mail weighs 1056 ounces. How much does that package weigh in pounds?

Mathematics

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

Step-by-step explanation:

16 oz equal a pound so 1056 divided by 16

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Today, the price of a new cell phone is $129. In 2000, the price of a similar cell phone was $240. What os the percent of change

9966 [12]

Answer:

Step-by-step explanation:

so first you look it up and if that dont work then u guess

8 0

3 years ago

A cat had a litter of 5 kittens. If each sex is equally likely, what is the probability that 3 or less were female kittens?

ASHA 777 [7]

Answer:

0.8125

Step-by-step explanation:

In this question, we are tasked with calculating the probability that 3 or less of her kittens were female.

Since each bsex is of likely probability, the probability of a male kitten = probability of a female kitten = 0.5

Now to calculate for 3 or less female kitten we are calcualting P(f) ≤ 3

In each case, we use the Bernoulli approximation

P(f) ≤ 3 = $5C3 m^{2} f^{3} + 5C2 m^{2} f^{3} + 5C1 m^{4} f + 5C0 m^{5} f^{0}$

Where m is the probability of a male kitten and f is the probability of having a female kitten with both values = 0.5

P(f) ≤ 3 =(0.3125) + (0.3125) + (0.15625) + (0.03125) = 0.8125

8 0

2 years ago

Area of octagon with radius 5?

Montano1993 [528]

I got 200 from a octagon with the diameter of 10

3 0

3 years ago

Teresa is simplifying an<br> Step 1: (4x²<br> Step 2:<br> Step 3: 2x

Lesechka [4]

The answer is B

None of the rest equal up to the same thing

7 0

3 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago