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Westkost [7]
3 years ago
5

Test worth 50 points; test score of 78%?

Mathematics
1 answer:
nataly862011 [7]3 years ago
7 0
The person would have scored 39 out of 50. Hope this helps
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Problem is in the attached picture. Thanks in advance!
bogdanovich [222]

Answer:

Width= 120 ft

Length= 285 ft

Step-by-step explanation:

W= width and (3w-75)= length since it's 75 less than 3 times the width

w+w+(3w-75)+(3w-75)=810

8w-150=810

8w=960

w=120

So the width is 120 ft

Then substitute the 120 in for w in the length

L= 3(120)-75

L=360-75

L=285

So the length is 285 ft

You can check because 120+120+285+285=810, and because 120×3-75=285!

6 0
4 years ago
a computer is on sale for $600 this is 75% of its original price. what was the original price of the computer?
MissTica
To get the answer we can use proportion
600 --------- 75%
x ------------- 100%
crossmultiply now
75x=600*100
75x=60000     /:75
x=800$ - its the answer
8 0
4 years ago
I need to answer A and B
ycow [4]
675840 feet that's the answer I'm sure of it
8 0
3 years ago
I'm taking a test and it's really important that I pass it because I'm already failing the class. Please help me I have no idea
soldi70 [24.7K]

Answer:

1. 15+5i

2. 3+4i

3. 3+i

4. 30

Step-by-step explanation:

4 0
3 years ago
In a random sample of cars driven at low altitudes, of them exceeded a standard of grams of particulate pollution per gallon of
Orlov [11]

Complete question is;

In a random sample of 370 cars driven at low altitudes, 43 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 80 cars driven at high altitudes, 23 of them exceeded the standard. Can you conclude that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard at an level of significance? Group of answer choices

Answer:

Yes we can conclude that there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard (P-value = 0.00005).

Step-by-step explanation:

This is a hypothesis test for the difference between the proportions.

The claim is that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

Then, the null and alternative hypothesis are:

H0 ; π1 - π2 = 0

H1 ; π1 - π2 < 0

The significance level would be established in 0.01.

The random sample 1 (low altitudes), of size n1 = 370 has a proportion of;

p1 = x1/n1

p1 = 43/370

p1 = 0.116

The random sample 2 (high altitudes), of size n2 = 80 has a proportion of;

p2 = x2/n2

p2 = 23/80

p2 = 0.288

The difference between proportions is pd = (p1-p2);

pd = p1 - p2 = 0.116 - 0.288

pd = -0.171

The pooled proportion, we need to calculate the standard error, is:

p = (x1 + x2)/(n1 + n2)

p = (43 + 23)/(370 + 80)

p = 66/450

p = 0.147

The estimated standard error of the difference between means is computed using the formula:

S_(p1-p2) = √[((p(1 - p)/n1) + ((p(1 - p)/n2)]

1 - p = 1 - 0.147 = 0.853

Thus;

S_(p1-p2) = √[((0.147 × 0.853)/370) + ((0.147 × 0.853)/80)]

S_(p1-p2) = 0.044

Now, we can use the formula for z-statistics as;

z = (pd - (π1 - π2))/S_(p1-p2)

z = (-0.171 - 0)/0.044

z = -3.89

Using z-distribution table, we have the p-value = 0.00005

Since the P-value of (0.00005) is smaller than the significance level (0.01), then the effect is significant.

We conclude that The null hypothesis is rejected.

Thus, there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

6 0
3 years ago
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