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allochka39001 [22]

3 years ago

13

Using coupons you spell $91 grocery shopping this is 65% of the total retail price of groceries how much would you have spent if

you had not use the coupons

1 answer:

Andrews [41]3 years ago

3 0

$140 because

$91 divided by 65 is 1.4
1.4 times 100 is 140

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The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 122 and standard deviation of 22

marin [14]

Answer: 4.27% of adults in the USA have stage 2 high blood pressure.

Step-by-step explanation:

Let x be a random variable that denotes a person with high blood pressure .

Given: Average blood pressure: $\mu=122$

Standard deviation: $\sigma=22$

Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher.

The probability that an adult in the USA have stage 2 high blood pressure:

$P(x\geq160)=P(\dfrac{x-\mu}{\sigma}}\geq\dfrac{160-122}{22})\\\\=P(z\geq1.72)\ \ \ [z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z$

Hence, 4.27% of adults in the USA have stage 2 high blood pressure.

6 0

3 years ago

What is the equation of the line that passes through (–2, –3) and is perpendicular to 2x – 3y = 6?

lyudmila [28]

<span>An equation that would be perpendicular is opposite the reciprocal. </span>

6 0

3 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

6 0

2 years ago

Read 2 more answers

What is the area of the shaded portion of the circle?

SOVA2 [1]

Answer:

The first option is the correct one, the area of the shaded portion of the circle is

[/tex](5 \pi -11.6)ft^2[/tex]

Step-by-step explanation:

Let us first consider the triangle + the shadow.

The full area of the circle is the radius squared times pi, so

A= $(5 ft)^2 \cdot \pi \\25 ft^2 \cdot \pi$

Since $\frac{72^{\circ}}{360^{\circ}}=\frac{1}{5}$ , the area of the triangle + the shaded area is one fifth of the area of the whole circle, thus

$A_1=\frac{1}{5}25 ft^2 \cdot \pi\\ =5 ft^2 \cdot \pi$

If we want to know the area of the shaded part of the circle, we must subtract the area of the triangle from $A_1$ .

The area of the triangle is given by

$A_{triangle}=\frac{1}{2}\cdot (2.9+2.9)ft \cdot 4 ft\\= 11.6 ft^2$

Thus the area of the shaded portion of the circle is

$A_1-A_{triangle}=5 \pi ft^2-11.6ft^2\\= (5 \pi -11.6)ft^2$

3 0

3 years ago

Bill needs to build a rectangular sheep pen

Ymorist [56]

Perimeter = 24m

Cost of half metre(1/2m) = €1.20

Cost of one metre = €2.40

Cost of fencing = 24 × 2.40 = €57.60

HOPE THIS WILL HELP YOU

7 0

4 years ago

Read 2 more answers