Ask question

Ask question

All categories

3 years ago

7

Linda has scored 84, 93, 91, and 60 on her previous four tests. What score does she need on her next test so that her average (m

ean) is 82?

1 answer:

vovikov84 [41]3 years ago

4 0

Answer:

She needs to score 82

Step-by-step explanation:

Mean of five test scores = 82

Let the score she got in fifth test = x

$\frac{84+93+91+60+x}{5}=82\\\\\frac{328+x}{5}=82\\\\328+x=82*5\\\\328+x=410\\\\x=410-328\\\\x=82$

You might be interested in

What comes next in the pattern?12,16,20<br>A. 10<br>B. 22<br>C. 23<br>D. 24

kolezko [41]

24 , it’s just skipping four letters up

7 0

3 years ago

. In a study about the relationship between the age of a person and the preference for digital vs analog watches, which of the f

Daniel [21]

Answer:

B and the first d and second one.

Step-by-step explanation:

The age is a big factor and b can showing the preference per age

6 0

3 years ago

Please help<br><br>p/5 = 3.1; p=

aalyn [17]

P=15.5 would be the answer

3 0

3 years ago

Read 2 more answers

A chemist has two alloys, one of which is 15% gold and 20% lead and the other which is 30% gold and 50% lead. How many grams of

vlabodo [156]

Answer: 490 grams of the first alloy should be used.

30 grams of the second alloy should be used.

Step-by-step explanation:

Let x represent the weight of the first alloy in grams that should be used.

Let y represent the weight of the second alloy in grams that should be used.

A chemist has two alloys, one of which is 15% gold and 20% lead. This means that the amount of gold and lead in the first alloy is

0.15x and 0.2x

The second alloy contains 30% gold and 50% lead. This means that the amount of gold and lead in the second alloy is

0.3y and 0.5y

If the alloy to be made contains 82.5 g of gold, it means that

0.15x + 0.3y = 82.5 - - - - - - - - - - - -1

The second alloy would also contain 113 g of lead. This means that

0.2x + 0.5y = 113 - - - - - - - - - - - - -2

Multiplying equation 1 by 0.2 and equation 2 by 0.15, it becomes

0.03x + 0.06y = 16.5

0.03x + 0.075y = 16.95

Subtracting, it becomes

- 0.015y = - 0.45

y = - 0.45/- 0.015

y = 30

Substituting y = 30 into equation 1, it becomes

0.15x + 0.3 × 30 = 82.5

0.15x + 9 = 82.5

0.15x = 82.5 - 9 = 73.5

x = 73.5/0.15

x = 490

6 0

3 years ago

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to in

oksano4ka [1.4K]

Answer:

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

$p_v =P(Z>2)=0.0228$

Step-by-step explanation:

1) Data given and notation

$\bar X=8300$ represent the sample mean

$\sigma=1200$ represent the population standard deviation

$n=64$ sample size

$\mu_o =800$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

4) P-value

Since is a one-side upper test the p value would given by:

$p_v =P(Z>2)=0.0228$

5) Conclusion

If we compare the p value and the significance level assumed, for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.

4 0

3 years ago