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3 years ago

What is the gradient of the blue line

Mathematics

1 answer:

katrin [286]3 years ago

5 0

Answer:The blue line is the tangent line that passes through the point (0.75, 0.25). The point on the tangent line is (3, 4). The gradient of the curve is approximately 0.6. Note: Parallel lines have the same gradient.

Step-by-step explanation:

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Seiji and Gavin both worked hard over the summer. Together they earned a total of $450. Gavin earned $30 more than seiji.

jok3333 [9.3K]

Answer:

(a) $s+g=450\\g=s+30$

(b) The graph is shown below.

(c) Seiji earned $ 210 and Gavin earned $ 240.

Step-by-step explanation:

Given:

Sum of earnings of Seiji and Gavin = $ 450

Gavin earns $ 30 more that that of Seiji.

(a)

So, as per question, if s and g are the amounts earned by Seiji and Gavin respectively, then;

$s+g=450\\g=s+30$

(b)

The graph is plotted using the x and y intercepts of each line.

For the line, $s+g=450$ , the x intercept is at $g=0$ . So, x intercept is (450,0). Similarly, the y intercept is when s=0, which is (0,450). Draw a line passing through these two points. The red line represents $s+g=450$ .

We follow the same process to plot the next line. The x intercept is at (-30,0) and y intercept is at (0,30). Draw a line passing through these two points. The blue line represents $g=s+30$ .

The graph is shown below.

(c)

From the graph, the point of intersection of the two lines gives the earnings of each of the persons.

The $x$ value of the point of intersection is the earning of Seiji which is $ 210 and the $y$ value is the earning of Gavin which is $ 240 as shown in the graph.

Therefore, Seiji earned $ 210 and Gavin earned $ 240.

3 0

3 years ago

Help me to solve this

antoniya [11.8K]

The answer is 2cm

$8 \times 10 {}^{ - 3} \times 250 = 2000 \times 10 {}^{ - 3} \\ = 2$
or:
$8 \times {10}^{ - 3} \times 250 = 0.008 \times 250 \\ = 2$

good luck

4 0

3 years ago

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many

kotykmax [81]

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

Answer:

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the first successful trial, then the probability of having k is given by

Pr(X=k) = $(1-p)^{k-1}p$

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

4 0

3 years ago

Plzzzzzz help!! I will give brainliest to whoever answers

jarptica [38.1K]

Answer:

4 and -3 i think

Step-by-step explanation:

the dot is on 4 and -3

5 0

3 years ago

Read 2 more answers

Can you guys help i’m kinda struggling

Brrunno [24]

Answer:

60 degrees

Step-by-step explanation:

equilateral triangles have all equal sides

6 0

3 years ago