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1 year ago

7

The line of best fit for a scatter plot is y=5.29x-219. The actual value of y when x is 57 is 90. What is the residual value whe

n x is 57?
A. -7.47
B. 82.53
C. 7.47
D. -33.00

1 answer:

Tems11 [23]1 year ago

7 0

The answer here is B.

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As the number of phones produced increases, wil the working rate ever get to 100%? Explain.

evablogger [386]

No, just because the phones produced Increases it doesn’t mean that the work rate will get up to 100%. There can always be more workers.

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2 years ago

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Stanley has a collection of seashells. He found 35% of his collection on Florida beaches. If Stanley has 49 seashells from Flori

m_a_m_a [10]

Answer:

The number of seashells he have in his collection all together is <u>140</u>.

Step-by-step explanation:

Given:

Stanley has a collection of seashells. He found 35% of his collection on Florida beaches.

Stanley has 49 seashells from Florida.

Now, to find the number of seashells of his collection altogether.

Let the number of seashells all together be $x.$

Percentage of seashells found on Florida beaches = 35%.

Number of seashells found on Florida beaches = 49.

Now, to get the number of seashells altogether we put an equation:

$35\%\ of\ x=49.$

⇒ $\frac{35}{100} \times x=49$

⇒ $0.35\times x=49$

⇒ $0.35x=49$

Dividing both sides by 0.35 we get:

⇒ $x=140.$

Therefore, the number of seashells he have in his collection all together is 140.

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1 year ago

A company logo is made up of a square and three identical triangles. What is the area of the logo? Enter your answer in the box.

garri49 [273]

Given:
square with sides measuring 7 cm.
3 triangles attached to three sides of the square. A line bisecting one triangle is measured at 4 cm.

Area of a square = s² = (7cm)² = 49 cm²

Area of a triangle = hb/2 = (4cm*7cm)/2 = 14 cm²
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Total area of the logo = 49 cm² + 42 cm² = 91 cm²

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Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly s

kifflom [539]

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.

Case 1: P(X=1) = [P (not infected)]⁵

= (0.15 - 0.1)⁵

P(X=1) = 3.125*10⁻⁷

Case 2: P(X=6) = 1- P(X=1)

= 1 - (1 - 0.1)⁵

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We can then use the previously determined values to compute the expected number of tests.

E(X) = ∑x.P(X=x)

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If they took the same route home itll be 13 miles

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