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netineya [11]
3 years ago
11

These 2 questions math 9th thanks! will give brainly!!

Mathematics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

1)     25+7(n-1)

When the above 'n'(hours) is substituted on the equation, it equals up to the right cost.

2)      Given that n represents the no.of bikes, you just plug it in to the equation below.

250+39(n-1)

= 250+39(20-1)

= 250+39*19

= 250+741

= 991

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2. What is the area of this triangle?
Brums [2.3K]

Answer:

I think it's answer is A. 6ft²

As Area = 1/2 *p*b

= 1/2 * 3*4

= 6 ft²

3 0
4 years ago
Please help. I’ll mark you as brainliest if correct!
Ivan

Answer:

A.

Step-by-step explanation:

5 0
4 years ago
How would you write 1,496,000,000 in scientific notation?
Verdich [7]
The answer is 1.496 times 10^9
8 0
4 years ago
Read 2 more answers
Which shows one way to determine the factors of x3 11x2 – 3x – 33 by grouping? x2(x 11) 3(x – 11) x2(x – 11) – 3(x – 11) x2(x 11
Aloiza [94]

The one way to determine factors of x³  + 11x² – 3x – 33  will be x^2(x+11)-3(x+11)_.

<h3>What is a factorization?</h3>

It is the method to separate the polynomial into parts and the parts will be in multiplication. And the value of the polynomial at this point will be zero.

The steps involved in the factorization are;

1. For each pair of parentheses, we create a common factor.

2. We use x2 as a common factor for the first parenthesis.

3. We use common factor 3 for the second parenthesis.

x^3  + 11x^2 - 3x -33 \\\\ x^2(x+11)-3(x+11)_

We will find the final solution as x^2(x+11)-3(x+11)_.

Hence the one way to determine factors of x³  + 11x² – 3x – 33  will be x^2(x+11)-3(x+11)_.

To learn more about the factorization refer to the link;

brainly.com/question/24182713

3 0
2 years ago
Philips Semiconductors is a leading European manufacturer of integrated circuits. Integrated circuits are produced on silicon wa
shepuryov [24]

Answer:

Step-by-step explanation:

1) The null hypothesis is,

H_0: The mean thickness of teh wafers for the five positions are equal

i.e, H_0:\mu_1=\mu_2=\mu_3=\mu_4=\mu_5

2)

The alternative hypothesis is,

H_1: There is an evidence of a difference in the mean thickness of the wafers for the five positions

3)

Let us consider the level of significance \alpha=0.01

from the Minitab outout

One-way ANOVA:C1 versus C2

source         DF            SS                 MS               F            P

C2                  4      1417.73          354.43        51.00       0.00

Error           145     1007.77              6.95

Total           14      2425.50

S = 2.636       R - S = 58.45%     R - Sq(adj) = 57.31%

Individual 95% CIs For Mean Based on Pooled StDev

level         N         Mean            StDev    -,----------,----------,----------,----------

1              30    240.53                2.62    (--,--)

2             30     243.73                2.79             (--,--)

3             30     246.07                2.90                           (--,--)

4             30     249.10                 2.66                                        (--,--)

5             30     247.07                 2.15                                  (--,--)

                                                               -,----------,----------,----------,----------

                                                        240.0   243.0   246.0   249.0

Pooled StDev = 2.64

4)

The test statistic is, F = 51

5)

The P-value is approximately 0

6)

Here, the P - value is less than the level of significance

\therefore \,P-value

So, we do not accept our null hypothesis H_0

7)

Therefore, we conclude that there is an evidence of a difference in the mean thickness of the wafers for the five positions at level of significance \alpha=0.05

b)

chek attachment

we observe that,

The mean thickness of the wafer for position 1 is significant with position 2,

position 18, position 19 and position 28.

The mean thickness of the wafer for position 2 is significant with position 18,

position 19 and position 28.

The mean thickness of the wafer for position 18 is significant with the

position 19.

But the mean thickness of the wafer for position18 is not significant with

position 28.

The mean thickness of the wafer for position 19 is significant with position 28.

7 0
3 years ago
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