What is the measure of ABC? A 44° B 88° C 176° D 92°

Carmen cuts a 60-inch-long wire into pieces that are 3/4 inch long.How many pieces does she have?

miskamm [114]

Since we have to divide 60 by ¾, so get 60/ ¾ =60* 4/3 = 80

7 0

3 years ago

John bought new headphones originally listed for $ 80.99 . They are \(35\%)| off. Which equation can be used to find the amount

Katarina [22]

Answer:

$28.35

Step-by-step explanation:

The headphones are $80.99 and are 35% off. To find the amount he will save, change the 35% to a decimal by moving it two times to the left, which is 0.35. Multiply 80.99 and 0.35 to get 28.3465. Round to get $28.35.

7 0

2 years ago

Shondra buys a bedroom suite on the in-stallment plan. She makes a down payment of $200 and monthly payments of $52 each for thr

Dmitry [639]

Answer:

$2072

Step-by-step explanation:

Note that:

1 year = 12 months

Hence:

1 year = 12 months

3 years = x months

Cross Multiply

x = 3 × 12

x = 36 months

She makes a down payment of $200 and monthly payments of $52 each for three years.

Her total cost is represented by the expression

= $200 + $52 × m

Where m = number of monthly payments.

$200 + $52 × 36

= $200 + $1872

= $ 2072

7 0

3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago

NEED HELP PLZ HURRY

Vlada [557]

(-2,8) (1,2)
m=2-8/1-(-2)
m=-6/3
m= -2
y=-2x+b
(-2,8)
8=-2(-2)+b
8=-4+b
12=b
b=12
y=-2x+12
Hope this helps!

6 0

3 years ago