Work out the value of the missing fraction

Build a fence around his rectangular garden. The length is 30 feet. The width is 6 yards. How many feet of fencing is needed

nadya68 [22]

The field that is being fenced in has two 30 foot sides and 2 18 foot sides. The reason they are 18 is because there are 3 feet in a yard. You would then add all 4 sides together, equaling your perimeter.

The final answer is 96 sq feet.

4 0

3 years ago

Select the expressions that are equivalent to 8(6m).

Bas_tet [7]

$\textsf{Hey there!}$

$\text{Select the expressions that are equicalent to 8(6m)}\downarrow$

$\text{We DON'T have any like terms so we CAN'T pair them off!}$

$\mathsf{8(6m) \rightarrow 8 \times 6 \times \ m}\\\mathsf{8\times 6 = 48}}\\\mathsf{48\times\ m = 48m}\\\\\\\boxed{\boxed{\mathsf{Answer: \bf{48m}}}}\checkmark$

$\textsf{Good luck on your assignment and enjoy your day!}$

~ $\dfrac{\frak{LoveYourselfFirst}}{:)}$

4 0

3 years ago

Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A samp

shutvik [7]

Answer:

0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .