All categories

3 years ago

I need help solving for this

Mathematics

2 answers:

puteri [66]3 years ago

8 0

I believe each teacher gets half a doughnut and mrs. Davis will have 5 doughnuts left

Ksenya-84 [330]3 years ago

7 0

Answer: 12/15 = 0.8

Step-by-step explanation: There are 12 donuts and 15 teachers, just divide the amount of donuts by the amount of teachers. Hope that helps.

You might be interested in

64,569 rounded to the thousands

Zina [86]

The 4 is in the thousands place and looking to the right of the 4 is a 5. The rule is less than 4 leave it the same; 5 or more add 1 to the number.

So the answer would be 65,000.

3 0

3 years ago

Find the value of x if 196^×=14

castortr0y [4]

Answer:

x = $\frac{1}{2}$

Step-by-step explanation:

Note that $\sqrt{196}$ = 14

Expressed in exponent form as

$196^{\frac{1}{2} }$ = 14, thus

x = $\frac{1}{2}$

7 0

3 years ago

CAN SOMEONE HELP ME FAST

a_sh-v [17]

Answer is <span>A. y=14x

hope that helps</span>

8 0

3 years ago

Read 2 more answers

Express (a+b) ^2 -4 by factorization

NemiM [27]

Step-by-step explanation:

$= (a + b) {}^{2} - 2 { }^{2} \\ = ( a + b + 2)(a + b - 2)$

in 2nd exp. this is because by using the formula of a2-b2

5 0

3 years ago

A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a samp

fenix001 [56]

Answer:

The lower bound of a 99% C.I for the proportion of defectives = 0.422

Step-by-step explanation:

From the given information:

The point estimate = sample proportion $\hat p$

$\hat p = \dfrac{x}{n}$

$\hat p = \dfrac{55}{100}$

$\hat p$ = 0.55

At Confidence interval of 99%, the level of significance = 1 - 0.99

= 0.01

$Z_{\alpha/2} =Z_{0.01/2} \\ \\ = Z_{0.005} = 2.576$

Then the margin of error $E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1-\hat p)}{n}}$

$E = 2.576 \times \sqrt{\dfrac{0.55(1-0.55)}{100}}$

$E = 2.576 \times \sqrt{\dfrac{0.2475}{100}}$

$E = 2.576 \times0.04975$

E = 0.128156

E ≅ 0.128

At 99% C.I for the population proportion p is: $\hat p - E$

= 0.55 - 0.128

= 0.422

Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422

6 0

3 years ago