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2 years ago

A package of 5 pairs of gloves costs $ 75.5 what is the cost of a single pair of gloves

Mathematics

1 answer:

zhenek [66]2 years ago

6 0

5 pairs of gloves, two gloves per pair, so we have 10 gloves.

If the entire pack cost $29.45, we divide that by 10 to determine the cost of a single glove:

$29.45 ÷ 10 = $2.945 which rounds to $2.95 per glove.

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PLEASE HELP ASAP <br> FIND THE MEASUREMENT QP<br> 20 POINTS

ahrayia [7]

Answer:

a^2+b^2=c^2

a^2+b^2=8^2

a^2+b^2=64

32+32=64

5.66^2+5.66^2=8^2

Step-by-step explanation:

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2 years ago

$500 shared in ratio 2:3:9

RideAnS [48]

2:3:9......added together = 14

2/14 (500) = 1000/14 = 71.43
3/14 (500) = 1500/14 = 107.14
9/14 (500) = 4500/14 = 321.43

6 0

3 years ago

Which number completes the system of linear inequalities<br> represented by the graph?

vaieri [72.5K]

Answer: What graph?

Step-by-step explanation:

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3 years ago

A car is traveling at 20.0 m/s, and the driver sees a traffic light. turn red. After 0.530 s (the reaction time), the driver app

slava [35]

Before the driver applies the brakes ( with the reaction time ):
d 1 = v0 · t = 20 m/s · 0.53 s = 10.6 m
After that:
v = v0 - a · t1
0 = 20 m/s - 7 · t1
7 · t1 = 20
t1 = 2.86 s
d 2 = v 0 · t1 - a · t1² / 2
d 2 = 20 m/s · 2.86 s - 7 m/s² · (2.86 s)²/2 = 57.2 m - 28.6 m = 28.6 m
d = d 1 + d 2 = 10.6 m + 28.6 m = 39.2 m
Answer: the stopping distance of a car is 39.2 m.

5 0

3 years ago

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal prob

Novay_Z [31]

Answer:

a) By the Central Limit Theorem, it is approximately normal.

b) The standard error of the distribution of the sample mean is 1.8333.

c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours

e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 36 hours and a standard deviation of 5.5 hours.

This means that $\mu = 36, \sigma = 5.5$

a. What can you say about the shape of the distribution of the sample mean?

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that $n = 9$ . So

$s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333$

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

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

By the Central Limit Theorem

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

$Z = \frac{38 - 36}{1.8333}$

$Z = 1.09$

$Z = 1.09$ has a pvalue of 0.8621

1 - 0.8621 = 0.1379

0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

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

$Z = \frac{34.5 - 36}{1.8333}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061.

1 - 0.2061 = 0.7939

0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.

e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?

pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

4 0

3 years ago