All categories

3 years ago

The original price of a computer was reduced by 120 dollars. The new,

Mathematics

1 answer:

wariber [46]3 years ago

4 0

Answer:

x-120=899

Step-by-step explanation:

(x=1019)

You might be interested in

Explain how you can use the graph of the quadratic function f(x) to solve the equation f(x)=0

Whitepunk [10]

F(x)=0
f(x)=y
so where y=0, the x value is the x in f(x)
basically, where the line crosses or touches the x axis, those ar the x values that make the function true

6 0

3 years ago

L has $200. Ty has 30% more than Lu and twice as much as Ali. How much money do they have altogether? (PLEASE SHOW STEPS!)

saul85 [17]

30% of 200=60 200+60= 260
Ty has £260
£260 x 2= £520

200+260+520= £980

Answer £980

I hope this is right :D

5 0

3 years ago

Help me with this math

stellarik [79]

Answer:

(x+2)(x+5) or if you were looking for the x-intercepts they would be -5, and -2

7 0

3 years ago

Read 2 more answers

8 is 4% of the number

Murljashka [212]

Answer:

(8/4)100 = 200

Step-by-step explanation:

Simplify the following:

8/4×100

Hint: | Express 8/4×100 as a single fraction.

8/4×100 = (8×100)/4:

(8×100)/4

Hint: | In (8×100)/4, divide 100 in the numerator by 4 in the denominator.

4 | | 2 | 5

| 1 | 0 | 0

- | | 8 |

| | 2 | 0

| - | 2 | 0

| | | 0:

8×25

Hint: | Multiply 8 and 25 together.

8×25 = 200:

Answer: 200

5 0

3 years ago

Read 2 more answers

An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing

wariber [46]

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 $n_1=15$

Spacing 1 $s_1=8$

Mean 1 $\=x_1=8.48$

Standard deviation 1 $\sigma_1=0.96 kPa$

Sample size 2 $n_2=16$

Spacing 2 $s_2=6$

Mean 1 $\=x_1=9.93$

Standard deviation 1 $\sigma_1=1.02 kPa$

Generally the hypothesis is mathematically given as

Null $H_0:\mu_1-\mu_2$

Alternative $H_a:\mu_1\mu_2$

Generally the equation for pooled estimate is mathematically given by

$S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }$

Therefore

$S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }$

$S=0.9905$

Generally the equation for test statistics is mathematically given by

$t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}$

$t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$t=-4.00091$

Therefore From table

$P value=P(t_{15+15-2}=4.00091)$

$P value=0.00041918$

$P$

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

3 0

3 years ago