All categories

3 years ago

1 QUESTION QUICKK EXPERTS/ACE OR ANYONE WHO WANTS TO HELP!!

Mathematics

1 answer:

solniwko [45]3 years ago

8 0

looks like all the statements are true,

Leighanne started at $15, they don't show a 0 for Martin

Martins 1st week is $65

leighanne, is saving 50 per week

leighanne is saving 50 per week, Martin is saving 40 per week

You might be interested in

Help ASAP! Correct answer will be marked BRAINLIEST!!!

Nataly_w [17]

Answer:

not telling u the answer but first multiply 2 x 2 1/2 (4 1/2) then add that to 6 2/5

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the answer please help!

Sav [38]

Answer:

(1,3)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

a cardboard box holds 31.25 cubic inches of creamy butter. The butter is molded into a square prism that is 5 inches long. How m

lesya [120]

Answer:

$62.5 \:in^2$

Step-by-step explanation:

Volume of the butter in the cardboard = 31.25 cubic inches

Length of the Molded Square Prism =5 Inches

Let the length of the square cross sectional area = x

Volume of the Prism $=x^2l$

Since l=5 Inches

$5x^2=31.25\\x^2=6.25\\x=2.5\: Inches$

Next, we determine the surface area of the New Cardboard box to be made.

Length of the Cardboard Box=5 Inches

Length of the square cross sectional area = 2.5 Inches

Total Surface Area $=2(x^2+xl+xl)=2(x^2+2xl)$

$=2(2.5^2+2(2.5)(5))\\=2(6.25+25)\\=2*31.25\\=62.5 \:in^2$

Therefore, $62.5 \:in^2$ of cardboard is needed,

4 0

3 years ago

Please help!

Aloiza [94]

Answer:

$4 \sin( \frac{2}{7}x ) + 1$

Step-by-step explanation:

Let use the basic function of sin first.

Since the midline is 1 and the max point is 5, the amplitude is 4.

The midline is 1 since the midline is at 0,1.

Since we know the distance of the max and midline x values

We can multiply that by 4 to find our period.

$\frac{7\pi}{4} \times 4 = 7\pi$

Use the equation 2pi/b to find the b we would use in our equation.

$\frac{2\pi}{b} = 7\pi$

The answer is

$\frac{2}{7}$

so our equation is

$4 \sin( \frac{2}{7}x ) + 1$

3 0

2 years ago

onsider the following hypothesis test: H 0: 50 H a: > 50 A sample of 50 is used and the population standard deviation is 6. U

kondaur [170]

Answer:

a) z(e) > z(c) 2.94 > 1.64 we are in the rejection zone for H₀ we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c) 1.18 < 1.64 we are in the acceptance region for H₀ we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12 > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point z(c) ; α = 0.05 fom z table w get z(c) = 1.64 we need to compare values (between z(c) and z(e) )

The test hypothesis is:

a) H₀ ⇒ μ₀ = 50 a) Hₐ μ > 50 ; for value 52.5

b) Hₐ μ > 50 ; for value 51

c) Hₐ μ > 50 ; for value 51.8

With value 52.5

The test statistic z(e) ??

a) z(e) = ( μ - μ₀ ) /( σ/√50) z(e) = (2.5*√50 )/6 z(e) = 2.94

2.94 > 1.64 we are in the rejected zone for H₀ we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) = ( μ - μ₀ ) /( σ/√50) ⇒ z(e) = √50/6 ⇒ z(e) = 1.18

z(e) < z(c) we are in the acceptance region for H₀ we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e) = ( μ - μ₀ ) /( σ/√50) ⇒ z(e) = (1.8*√50)/ 6 ⇒ z(e) = 2.12

2.12 > 1.64 and we can conclude the same as in case a

8 0

3 years ago