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2 years ago

Aliya buys p pounds of apples for $2 per pounds. Write an algebraic expression to find the total cost?

Mathematics

1 answer:

Burka [1]2 years ago

3 0

Answer:

Identify the terms and coefficients of each expression. 1. ... Frank buys p pounds of oranges for $2.29 per pound and the same number of pounds of apples for $1.69 per pound. ... .690 = total cost ... Write an expression to represent each situation. 9. Eliza earns $400 per week plus $15 for each new customer she signs up.

Step-by-step explanation:

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The straight line 2x + 3y = 7 meets the curve

NISA [10]

Step-by-step explanation:

xy + 10 = 0. => y = -10/x.

2x + 3y = 7. => 3y = -2x + 7, y = -2/3 x + 7/3.

When -10/x = -2/3 x + 7/3,

-10 = -2/3 x² + 7/3 x, 2/3 x² - 7/3 x - 10 = 0.

=> 2x² - 7x - 30 = 0

=> (2x + 5)(x - 6) = 0

=> x = -2.5 or x = 6.

dy/dx = d/dx [-10/x] = 10/x².

When x = -2.5, dy/dx = 10 / (-2.5)² = 1.6.

When x = 6, dy/dx = 10 / (6)² = 5/18.

Hence the gradients at the points are 1.6 and 5/18.

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2 years ago

Lines s, t, u, and v are cut by transversal r. Clockwise from top left, the angles formed by r and s are blank, 89 degrees, blan

Natali [406]

Answer:

its not b but i think its c

Step-by-step explanation:

because it is

8 0

3 years ago

Read 2 more answers

What is the vertex form of the function y = x2 + 8x – 3

Scrat [10]

Answer:

y=(x+4)²-19

Step-by-step explanation:

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3 years ago

A survey of 700 adults from a certain region asked, "What do you buy from your mobile device?" The results indicated that 59%

Kryger [21]

Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes $n_{1} = 400 and n_{2} = 300$

Proportion of mean $p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52$

Null hypothesis H0 : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

Alternative hypothesis H1:- p1 ≠ p2

a) The test statistic is

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }$

where $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

on calculation we get p = 0.56

now q =1-p = 1-0.56=0.44

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }\\ =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300} }$

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

conclusion:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) 95% confidence intervals

The confidence intervals are P± 1.96(√PQ/n)

we know that = $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of confidence intervals (0.523 ,0.596)

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3 years ago

The student council at woodland middle school is planning a field trip. They polled randomly selected teachers and students and

ElenaW [278]

19 percent of the people polled who preferred burgers were teachers.

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3 years ago

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