Whats the answer then

Can someone please answer this

kaheart [24]

Answer:

<h2> ${7}^{3} (as \: a \: single \: power \: of \: 7)$ </h2>

Step-by-step explanation:

Given 

${7}^{4} \div 7$

Since,

${a}^{x} \div {a}^{y} = {a}^{x - y}$

Hence,

$= {7}^{4 - 3}$

$= {7}^{3} (ans)(as \: a \: single \: power \: of \: 7)$

8 0

2 years ago

The top of a ladder rests at a height of 15 feet against the side of a house. If the base of the ladder is 6 feet from the house

Ksju [112]

90
15 x 6 = 90
You need to multiply

6 0

3 years ago

Read 2 more answers

How do you solve |x-4| is less than or equal to 2?

Harlamova29_29 [7]

Answer:

with your brain

Step-by-step explanation:

duhh8/7

7 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$